我正在尝试编写一款可以帮助您管理财务状况的应用。我正在使用EditText字段，用户可以在其中指定金额。
我将inputType设置为numberDecimal，但工作正常，但这样可以让人们输入123.122等数字并不适合赚钱。
有没有办法将小数点后的字符数限制为两个？
答案 0 :(得分：102)
更优雅的方式是使用正则表达式（正则表达式），如下所示：
public class DecimalDigitsInputFilter implements InputFilter {
Pattern mPattern;
public DecimalDigitsInputFilter(int digitsBeforeZero,int digitsAfterZero) {
mPattern=Pattern.compile("[0-9]{0," + (digitsBeforeZero-1) + "}+((\\.[0-9]{0," + (digitsAfterZero-1) + "})?)||(\\.)?");
}
@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
Matcher matcher=mPattern.matcher(dest);
if(!matcher.matches())
return "";
return null;
}
}
要使用它，请执行以下操作：
editText.setFilters(new InputFilter[] {new DecimalDigitsInputFilter(5,2)});
答案 1 :(得分：58)
不使用正则表达式的简单解决方案：
import android.text.InputFilter;
import android.text.Spanned;
/**
* Input filter that limits the number of decimal digits that are allowed to be
* entered.
*/
public class DecimalDigitsInputFilter implements InputFilter {
private final int decimalDigits;
/**
* Constructor.
*
* @param decimalDigits maximum decimal digits
*/
public DecimalDigitsInputFilter(int decimalDigits) {
this.decimalDigits = decimalDigits;
}
@Override
public CharSequence filter(CharSequence source,
int start,
int end,
Spanned dest,
int dstart,
int dend) {
int dotPos = -1;
int len = dest.length();
for (int i = 0; i < len; i++) {
char c = dest.charAt(i);
if (c == '.' || c == ',') {
dotPos = i;
break;
}
}
if (dotPos >= 0) {
// protects against many dots
if (source.equals(".") || source.equals(","))
{
return "";
}
// if the text is entered before the dot
if (dend <= dotPos) {
return null;
}
if (len - dotPos > decimalDigits) {
return "";
}
}
return null;
}
}
使用：
editText.setFilters(new InputFilter[] {new DecimalDigitsInputFilter(2)});
答案 2 :(得分：31)
InputFilter的这种实现解决了这个问题。
import android.text.SpannableStringBuilder;
import android.text.Spanned;
import android.text.method.DigitsKeyListener;
public class MoneyValueFilter extends DigitsKeyListener {
public MoneyValueFilter() {
super(false, true);
}
private int digits = 2;
public void setDigits(int d) {
digits = d;
}
@Override
public CharSequence filter(CharSequence source, int start, int end,
Spanned dest, int dstart, int dend) {
CharSequence out = super.filter(source, start, end, dest, dstart, dend);
// if changed, replace the source
if (out != null) {
source = out;
start = 0;
end = out.length();
}
int len = end - start;
// if deleting, source is empty
// and deleting can't break anything
if (len == 0) {
return source;
}
int dlen = dest.length();
// Find the position of the decimal .
for (int i = 0; i < dstart; i++) {
if (dest.charAt(i) == '.') {
// being here means, that a number has
// been inserted after the dot
// check if the amount of digits is right
return (dlen-(i+1) + len > digits) ?
"" :
new SpannableStringBuilder(source, start, end);
}
}
for (int i = start; i < end; ++i) {
if (source.charAt(i) == '.') {
// being here means, dot has been inserted
// check if the amount of digits is right
if ((dlen-dend) + (end-(i + 1)) > digits)
return "";
else
break; // return new SpannableStringBuilder(source, start, end);
}
}
// if the dot is after the inserted part,
// nothing can break
return new SpannableStringBuilder(source, start, end);
}
}
答案 3 :(得分：29)
以下是 InputFilter 示例，其中只允许小数点前最多4位数，之后最多1位数。
edittext允许的值： 555.2 ， 555 ， .2
编辑文字块的值： 55555.2 ， 055.2 ， 555.42
InputFilter filter = new InputFilter() {
final int maxDigitsBeforeDecimalPoint=4;
final int maxDigitsAfterDecimalPoint=1;
@Override
public CharSequence filter(CharSequence source, int start, int end,
Spanned dest, int dstart, int dend) {
StringBuilder builder = new StringBuilder(dest);
builder.replace(dstart, dend, source
.subSequence(start, end).toString());
if (!builder.toString().matches(
"(([1-9]{1})([0-9]{0,"+(maxDigitsBeforeDecimalPoint-1)+"})?)?(\\.[0-9]{0,"+maxDigitsAfterDecimalPoint+"})?"
)) {
if(source.length()==0)
return dest.subSequence(dstart, dend);
return "";
}
return null;
}
};
mEdittext.setFilters(new InputFilter[] { filter });
答案 4 :(得分：18)
我为@Pinhassi解决方案做了一些修复。它处理一些情况：
1.您可以将光标移动到任何地方
2.minus sign handling
3.digitsbefore = 2且digitsafter = 4，您输入12.4545。然后，如果你想删除＆＃34;。＆＃34;，它将不允许。
public class DecimalDigitsInputFilter implements InputFilter {
private int mDigitsBeforeZero;
private int mDigitsAfterZero;
private Pattern mPattern;
private static final int DIGITS_BEFORE_ZERO_DEFAULT = 100;
private static final int DIGITS_AFTER_ZERO_DEFAULT = 100;
public DecimalDigitsInputFilter(Integer digitsBeforeZero, Integer digitsAfterZero) {
this.mDigitsBeforeZero = (digitsBeforeZero != null ? digitsBeforeZero : DIGITS_BEFORE_ZERO_DEFAULT);
this.mDigitsAfterZero = (digitsAfterZero != null ? digitsAfterZero : DIGITS_AFTER_ZERO_DEFAULT);
mPattern = Pattern.compile("-?[0-9]{0," + (mDigitsBeforeZero) + "}+((\\.[0-9]{0," + (mDigitsAfterZero)
+ "})?)||(\\.)?");
}
@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
String replacement = source.subSequence(start, end).toString();
String newVal = dest.subSequence(0, dstart).toString() + replacement
+ dest.subSequence(dend, dest.length()).toString();
Matcher matcher = mPattern.matcher(newVal);
if (matcher.matches())
return null;
if (TextUtils.isEmpty(source))
return dest.subSequence(dstart, dend);
else
return "";
}
}
答案 5 :(得分：17)
我通过以下方式在TextWatcher的帮助下实现了这一目标
final EditText et = (EditText) findViewById(R.id.EditText1);
et.addTextChangedListener(new TextWatcher() {
public void onTextChanged(CharSequence arg0, int arg1, int arg2,int arg3) {
}
public void beforeTextChanged(CharSequence arg0, int arg1,int arg2, int arg3) {
}
public void afterTextChanged(Editable arg0) {
if (arg0.length() > 0) {
String str = et.getText().toString();
et.setOnKeyListener(new OnKeyListener() {
public boolean onKey(View v, int keyCode, KeyEvent event) {
if (keyCode == KeyEvent.KEYCODE_DEL) {
count--;
InputFilter[] fArray = new InputFilter[1];
fArray[0] = new InputFilter.LengthFilter(100);
et.setFilters(fArray);
//change the edittext's maximum length to 100.
//If we didn't change this the edittext's maximum length will
//be number of digits we previously entered.
}
return false;
}
});
char t = str.charAt(arg0.length() - 1);
if (t == '.') {
count = 0;
}
if (count >= 0) {
if (count == 2) {
InputFilter[] fArray = new InputFilter[1];
fArray[0] = new InputFilter.LengthFilter(arg0.length());
et.setFilters(fArray);
//prevent the edittext from accessing digits
//by setting maximum length as total number of digits we typed till now.
}
count++;
}
}
}
});
此解决方案不允许用户在小数点后输入两位以上的数字。您也可以在小数点前输入任意位数。请参阅此博客http://v4all123.blogspot.com/2013/05/set-limit-for-fraction-in-decimal.html，为多个EditText设置过滤器。我希望这将有所帮助。谢谢。
答案 6 :(得分：11)
我的解决方案很简单，效果很好！
public class DecimalInputTextWatcher implements TextWatcher {
private String mPreviousValue;
private int mCursorPosition;
private boolean mRestoringPreviousValueFlag;
private int mDigitsAfterZero;
private EditText mEditText;
public DecimalInputTextWatcher(EditText editText, int digitsAfterZero) {
mDigitsAfterZero = digitsAfterZero;
mEditText = editText;
mPreviousValue = "";
mRestoringPreviousValueFlag = false;
}
@Override
public void beforeTextChanged(CharSequence s, int start, int count, int after) {
if (!mRestoringPreviousValueFlag) {
mPreviousValue = s.toString();
mCursorPosition = mEditText.getSelectionStart();
}
}
@Override
public void onTextChanged(CharSequence s, int start, int before, int count) {
}
@Override
public void afterTextChanged(Editable s) {
if (!mRestoringPreviousValueFlag) {
if (!isValid(s.toString())) {
mRestoringPreviousValueFlag = true;
restorePreviousValue();
}
} else {
mRestoringPreviousValueFlag = false;
}
}
private void restorePreviousValue() {
mEditText.setText(mPreviousValue);
mEditText.setSelection(mCursorPosition);
}
private boolean isValid(String s) {
Pattern patternWithDot = Pattern.compile("[0-9]*((\\.[0-9]{0," + mDigitsAfterZero + "})?)||(\\.)?");
Pattern patternWithComma = Pattern.compile("[0-9]*((,[0-9]{0," + mDigitsAfterZero + "})?)||(,)?");
Matcher matcherDot = patternWithDot.matcher(s);
Matcher matcherComa = patternWithComma.matcher(s);
return matcherDot.matches() || matcherComa.matches();
}
}
用法：
myTextEdit.addTextChangedListener(new DecimalInputTextWatcher(myTextEdit, 2));
答案 7 :(得分：8)
我不喜欢其他解决方案而且我创建了自己的解决方案。 使用此解决方案，您无法在该点之前输入超过MAX_BEFORE_POINT的数字，并且小数数字不能超过MAX_DECIMAL。
你无法输入过多的数字，没有其他效果！ 另外如果你写＆＃34;。＆＃34;它键入＆＃34; 0。＆＃34;
将布局中的EditText设置为：
机器人：的inputType =＆＃34; numberDecimal＆＃34;
在onCreate中添加监听器。如果你想修改点之前和之后的数字编辑调用PerfectDecimal（str，NUMBER_BEFORE_POINT，NUMBER_DECIMALS），这里设置为3和2
EditText targetEditText = (EditText)findViewById(R.id.targetEditTextLayoutId);
targetEditText.addTextChangedListener(new TextWatcher() {
public void onTextChanged(CharSequence arg0, int arg1, int arg2, int arg3) {}
public void beforeTextChanged(CharSequence arg0, int arg1, int arg2, int arg3) {}
public void afterTextChanged(Editable arg0) {
String str = targetEditText.getText().toString();
if (str.isEmpty()) return;
String str2 = PerfectDecimal(str, 3, 2);
if (!str2.equals(str)) {
targetEditText.setText(str2);
int pos = targetEditText.getText().length();
targetEditText.setSelection(pos);
}
}
});
包括这个Funcion：
public String PerfectDecimal(String str, int MAX_BEFORE_POINT, int MAX_DECIMAL){
if(str.charAt(0) == '.') str = "0"+str;
int max = str.length();
String rFinal = "";
boolean after = false;
int i = 0, up = 0, decimal = 0; char t;
while(i < max){
t = str.charAt(i);
if(t != '.' && after == false){
up++;
if(up > MAX_BEFORE_POINT) return rFinal;
}else if(t == '.'){
after = true;
}else{
decimal++;
if(decimal > MAX_DECIMAL)
return rFinal;
}
rFinal = rFinal + t;
i++;
}return rFinal;
}
已经完成了！
答案 8 :(得分：8)
小数点后要求 2位数。小数点前的数字应该有无限制。所以，解决方案应该是，
public class DecimalDigitsInputFilter implements InputFilter {
Pattern mPattern;
public DecimalDigitsInputFilter() {
mPattern = Pattern.compile("[0-9]*+((\\.[0-9]?)?)||(\\.)?");
}
@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
Matcher matcher = mPattern.matcher(dest);
if (!matcher.matches())
return "";
return null;
}
}
并将其用作，
mEditText.setFilters(new InputFilter[]{new DecimalDigitsInputFilter()});
感谢@Pinhassi的灵感。
答案 9 :(得分：7)
我想出的InputFilter允许您配置小数位前后的位数。此外，它不允许前导零。
public class DecimalDigitsInputFilter implements InputFilter
{
Pattern pattern;
public DecimalDigitsInputFilter(int digitsBeforeDecimal, int digitsAfterDecimal)
{
pattern = Pattern.compile("(([1-9]{1}[0-9]{0," + (digitsBeforeDecimal - 1) + "})?||[0]{1})((\\.[0-9]{0," + digitsAfterDecimal + "})?)||(\\.)?");
}
@Override public CharSequence filter(CharSequence source, int sourceStart, int sourceEnd, Spanned destination, int destinationStart, int destinationEnd)
{
// Remove the string out of destination that is to be replaced.
String newString = destination.toString().substring(0, destinationStart) + destination.toString().substring(destinationEnd, destination.toString().length());
// Add the new string in.
newString = newString.substring(0, destinationStart) + source.toString() + newString.substring(destinationStart, newString.length());
// Now check if the new string is valid.
Matcher matcher = pattern.matcher(newString);
if(matcher.matches())
{
// Returning null indicates that the input is valid.
return null;
}
// Returning the empty string indicates the input is invalid.
return "";
}
}
// To use this InputFilter, attach it to your EditText like so:
final EditText editText = (EditText) findViewById(R.id.editText);
EditText.setFilters(new InputFilter[]{new DecimalDigitsInputFilter(4, 4)});
答案 10 :(得分：5)
稍微改进了@Pinhassi解决方案。
效果很好。它验证连接的字符串。
public class DecimalDigitsInputFilter implements InputFilter {
Pattern mPattern;
public DecimalDigitsInputFilter() {
mPattern = Pattern.compile("([1-9]{1}[0-9]{0,2}([0-9]{3})*(\\.[0-9]{0,2})?|[1-9]{1}[0-9]{0,}(\\.[0-9]{0,2})?|0(\\.[0-9]{0,2})?|(\\.[0-9]{1,2})?)");
}
@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
String formatedSource = source.subSequence(start, end).toString();
String destPrefix = dest.subSequence(0, dstart).toString();
String destSuffix = dest.subSequence(dend, dest.length()).toString();
String result = destPrefix + formatedSource + destSuffix;
result = result.replace(",", ".");
Matcher matcher = mPattern.matcher(result);
if (matcher.matches()) {
return null;
}
return "";
}
}
答案 11 :(得分：5)
在将字符串放入TextView之前，请尝试使用NumberFormat.getCurrencyInstance()格式化字符串。
类似的东西：
NumberFormat currency = NumberFormat.getCurrencyInstance();
myTextView.setText(currency.format(dollars));
修改 - 我没有在文档中找到的货币输入类型。我想这是因为有些货币不遵循相同的小数位规则，例如日元。
正如LeffelMania所提到的，您可以使用TextWatcher上设置的EditText使用上述代码来更正用户输入。
答案 12 :(得分：5)
我修改了上述解决方案并创建了以下解决方案。您可以设置小数点前后的位数。
Namespace prefix mlhim2 on pcs-a768a3d0-c7e8-435d-b1db-cd7211aba6bc is not defined
}
答案 13 :(得分：3)
我改进了Pinhassi使用正则表达式的解决方案，因此它也正确处理边缘情况。在检查输入是否正确之前，首先构建最终字符串，如android docs所述。
public class DecimalDigitsInputFilter implements InputFilter {
private Pattern mPattern;
private static final Pattern mFormatPattern = Pattern.compile("\\d+\\.\\d+");
public DecimalDigitsInputFilter(int digitsBeforeDecimal, int digitsAfterDecimal) {
mPattern = Pattern.compile(
"^\\d{0," + digitsBeforeDecimal + "}([\\.,](\\d{0," + digitsAfterDecimal +
"})?)?$");
}
@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest,
int dstart, int dend) {
String newString =
dest.toString().substring(0, dstart) + source.toString().substring(start, end)
+ dest.toString().substring(dend, dest.toString().length());
Matcher matcher = mPattern.matcher(newString);
if (!matcher.matches()) {
return "";
}
return null;
}
}
用法：
editText.setFilters(new InputFilter[] {new DecimalDigitsInputFilter(5,2)});
答案 14 :(得分：3)
我已经改变了答案№6（由Favas Kv），因为那里你可以把点放在第一个位置。
final InputFilter [] filter = { new InputFilter() {
@Override
public CharSequence filter(CharSequence source, int start, int end,
Spanned dest, int dstart, int dend) {
StringBuilder builder = new StringBuilder(dest);
builder.replace(dstart, dend, source
.subSequence(start, end).toString());
if (!builder.toString().matches(
"(([1-9]{1})([0-9]{0,4})?(\\.)?)?([0-9]{0,2})?"
)) {
if(source.length()==0)
return dest.subSequence(dstart, dend);
return "";
}
return null;
}
}};
答案 15 :(得分：3)
Simple Helper类用于防止用户在十进制后输入超过2位的数字：
public class CostFormatter implements TextWatcher {
private final EditText costEditText;
public CostFormatter(EditText costEditText) {
this.costEditText = costEditText;
}
@Override
public void beforeTextChanged(CharSequence s, int start, int count, int after) {
}
@Override
public void onTextChanged(CharSequence s, int start, int before, int count) {
}
@Override
public synchronized void afterTextChanged(final Editable text) {
String cost = text.toString().trim();
if(!cost.endsWith(".") && cost.contains(".")){
String numberBeforeDecimal = cost.split("\\.")[0];
String numberAfterDecimal = cost.split("\\.")[1];
if(numberAfterDecimal.length() > 2){
numberAfterDecimal = numberAfterDecimal.substring(0, 2);
}
cost = numberBeforeDecimal + "." + numberAfterDecimal;
}
costEditText.removeTextChangedListener(this);
costEditText.setText(cost);
costEditText.setSelection(costEditText.getText().toString().trim().length());
costEditText.addTextChangedListener(this);
}
}
答案 16 :(得分：3)
这里的所有答案都非常复杂我试图让它变得更简单。看看我的代码并自己决定 -
int temp = 0;
int check = 0;
editText.addTextChangedListener(new TextWatcher() {
@Override
public void onTextChanged(CharSequence s, int start, int before, int count) {
if(editText.getText().toString().length()<temp)
{
if(!editText.getText().toString().contains("."))
editText.setFilters(new InputFilter[] { new InputFilter.LengthFilter(editText.getText().toString().length()-1) });
else
editText.setFilters(new InputFilter[] { new InputFilter.LengthFilter(editText.getText().toString().length()+1) });
}
if(!editText.getText().toString().contains("."))
{
editText.setFilters(new InputFilter[] { new InputFilter.LengthFilter(editText.getText().toString().length()+1) });
check=0;
}
else if(check==0)
{
check=1;
editText.setFilters(new InputFilter[] { new InputFilter.LengthFilter(editText.getText().toString().length()+2) });
}
}
@Override
public void beforeTextChanged(CharSequence s, int start, int count,
int after) {
temp = editText.getText().toString().length();
}
@Override
public void afterTextChanged(Editable s) {
// TODO Auto-generated method stub
}
});
答案 17 :(得分：3)
DecimalFormat form = new DecimalFormat("#.##", new DecimalFormatSymbols(Locale.US));
EditText et;
et.setOnEditorActionListener(new TextView.OnEditorActionListener() {
@Override
public boolean onEditorAction(TextView v, int actionId, KeyEvent event) {
if (actionId == EditorInfo.IME_ACTION_DONE) {
double a = Double.parseDouble(et.getText().toString());
et.setText(form.format(a));
}
return false;
}
});
这样做是当你退出编辑阶段时它将字段格式化为正确的格式。在他们的时刻，它只有2个小数字符号。我认为这是非常简单的方法。
答案 18 :(得分：2)
我真的很喜欢Pinhassi的回答，但注意到在用户输入小数点后的指定数字后，您无法再在小数点左侧输入文本。问题是该解决方案仅测试了之前输入的文本，而不是当前输入的文本。所以这是我的解决方案，将新字符插入到原始文本中进行验证。
package com.test.test;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
import android.text.InputFilter;
import android.text.Spanned;
import android.util.Log;
public class InputFilterCurrency implements InputFilter {
Pattern moPattern;
public InputFilterCurrency(int aiMinorUnits) {
// http://www.regexplanet.com/advanced/java/index.html
moPattern=Pattern.compile("[0-9]*+((\\.[0-9]{0,"+ aiMinorUnits + "})?)||(\\.)?");
} // InputFilterCurrency
@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
String lsStart = "";
String lsInsert = "";
String lsEnd = "";
String lsText = "";
Log.d("debug", moPattern.toString());
Log.d("debug", "source: " + source + ", start: " + start + ", end:" + end + ", dest: " + dest + ", dstart: " + dstart + ", dend: " + dend );
lsText = dest.toString();
// If the length is greater then 0, then insert the new character
// into the original text for validation
if (lsText.length() > 0) {
lsStart = lsText.substring(0, dstart);
Log.d("debug", "lsStart : " + lsStart);
// Check to see if they have deleted a character
if (source != "") {
lsInsert = source.toString();
Log.d("debug", "lsInsert: " + lsInsert);
} // if
lsEnd = lsText.substring(dend);
Log.d("debug", "lsEnd : " + lsEnd);
lsText = lsStart + lsInsert + lsEnd;
Log.d("debug", "lsText : " + lsText);
} // if
Matcher loMatcher = moPattern.matcher(lsText);
Log.d("debug", "loMatcher.matches(): " + loMatcher.matches() + ", lsText: " + lsText);
if(!loMatcher.matches()) {
return "";
}
return null;
} // CharSequence
} // InputFilterCurrency
设置editText过滤器的调用
editText.setFilters(new InputFilter[] {new InputFilterCurrency(2)});
Ouput with two decimal places
05-22 15:25:33.434: D/debug(30524): [0-9]*+((\.[0-9]{0,2})?)||(\.)?
05-22 15:25:33.434: D/debug(30524): source: 5, start: 0, end:1, dest: 123.4, dstart: 5, dend: 5
05-22 15:25:33.434: D/debug(30524): lsStart : 123.4
05-22 15:25:33.434: D/debug(30524): lsInsert: 5
05-22 15:25:33.434: D/debug(30524): lsEnd :
05-22 15:25:33.434: D/debug(30524): lsText : 123.45
05-22 15:25:33.434: D/debug(30524): loMatcher.matches(): true, lsText: 123.45
Ouput inserting a 5 in the middle
05-22 15:26:17.624: D/debug(30524): [0-9]*+((\.[0-9]{0,2})?)||(\.)?
05-22 15:26:17.624: D/debug(30524): source: 5, start: 0, end:1, dest: 123.45, dstart: 2, dend: 2
05-22 15:26:17.624: D/debug(30524): lsStart : 12
05-22 15:26:17.624: D/debug(30524): lsInsert: 5
05-22 15:26:17.624: D/debug(30524): lsEnd : 3.45
05-22 15:26:17.624: D/debug(30524): lsText : 1253.45
05-22 15:26:17.624: D/debug(30524): loMatcher.matches(): true, lsText: 1253.45
答案 19 :(得分：2)
像其他人说的那样，我在我的项目中添加了这个类，并将过滤器设置为我想要的EditText。
过滤器是从@ Pixel的答案中复制的。我只是把它们放在一起。
public class DecimalDigitsInputFilter implements InputFilter {
Pattern mPattern;
public DecimalDigitsInputFilter() {
mPattern = Pattern.compile("([1-9]{1}[0-9]{0,2}([0-9]{3})*(\\.[0-9]{0,2})?|[1-9]{1}[0-9]{0,}(\\.[0-9]{0,2})?|0(\\.[0-9]{0,2})?|(\\.[0-9]{1,2})?)");
}
@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
String formatedSource = source.subSequence(start, end).toString();
String destPrefix = dest.subSequence(0, dstart).toString();
String destSuffix = dest.subSequence(dend, dest.length()).toString();
String result = destPrefix + formatedSource + destSuffix;
result = result.replace(",", ".");
Matcher matcher = mPattern.matcher(result);
if (matcher.matches()) {
return null;
}
return "";
}
}
现在在您的EditText中设置过滤器。
mEditText.setFilters(new InputFilter[]{new DecimalDigitsInputFilter()});
这里有一件重要的事情是它确实解决了我在EditText小数点后不允许显示两位数以上的问题，但问题是来自getText()的{{1}} ，它返回我输入的整个输入。
例如，在EditText上应用过滤器后，我尝试设置输入1.5699856987。所以在屏幕上显示1.56是完美的。
然后我想将此输入用于其他一些计算，因此我想从该输入字段（EditText）获取文本。当我打电话给EditText时，它会返回1.5699856987，这在我的情况下是不可接受的。
因此我必须在从mEditText.getText().toString()获取值后再次解析该值。
EditTextBigDecimal amount = new BigDecimal(Double.parseDouble(mEditText.getText().toString().trim()))
.setScale(2, RoundingMode.HALF_UP);
获取完整文本后， setScale可以解决问题。
答案 20 :(得分：1)
@ Meh for u ..
txtlist.setFilters(new InputFilter[] { new DigitsKeyListener( Boolean.FALSE,Boolean.TRUE) {
int beforeDecimal = 7;
int afterDecimal = 2;
@Override
public CharSequence filter(CharSequence source, int start, int end,Spanned dest, int dstart, int dend) {
String etText = txtlist.getText().toString();
String temp = txtlist.getText() + source.toString();
if (temp.equals(".")) {
return "0.";
} else if (temp.toString().indexOf(".") == -1) {
// no decimal point placed yet
if (temp.length() > beforeDecimal) {
return "";
}
} else {
int dotPosition ;
int cursorPositon = txtlistprice.getSelectionStart();
if (etText.indexOf(".") == -1) {
dotPosition = temp.indexOf(".");
}else{
dotPosition = etText.indexOf(".");
}
if(cursorPositon <= dotPosition){
String beforeDot = etText.substring(0, dotPosition);
if(beforeDot.length()<beforeDecimal){
return source;
}else{
if(source.toString().equalsIgnoreCase(".")){
return source;
}else{
return "";
}
}
}else{
temp = temp.substring(temp.indexOf(".") + 1);
if (temp.length() > afterDecimal) {
return "";
}
}
}
return super.filter(source, start, end, dest, dstart, dend);
}
} });
答案 21 :(得分：1)
此代码效果很好
public class DecimalDigitsInputFilter implements InputFilter {
private final int digitsBeforeZero;
private final int digitsAfterZero;
private Pattern mPattern;
public DecimalDigitsInputFilter(int digitsBeforeZero, int digitsAfterZero) {
this.digitsBeforeZero = digitsBeforeZero;
this.digitsAfterZero = digitsAfterZero;
applyPattern(digitsBeforeZero, digitsAfterZero);
}
private void applyPattern(int digitsBeforeZero, int digitsAfterZero) {
mPattern = Pattern.compile("[0-9]{0," + (digitsBeforeZero - 1) + "}+((\\.[0-9]{0," + (digitsAfterZero - 1) + "})?)|(\\.)?");
}
@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
if (dest.toString().contains(".") || source.toString().contains("."))
applyPattern(digitsBeforeZero + 2, digitsAfterZero);
else
applyPattern(digitsBeforeZero, digitsAfterZero);
Matcher matcher = mPattern.matcher(dest);
if (!matcher.matches())
return "";
return null;
}
}
应用过滤器：
edittext.setFilters(new InputFilter[] {new DecimalDigitsInputFilter(5,2)});
答案 22 :(得分：1)
我也遇到过这个问题。我希望能够在许多EditTexts中重用代码。这是我的解决方案：
用法：
CurrencyFormat watcher = new CurrencyFormat();
priceEditText.addTextChangedListener(watcher);
类别：
public static class CurrencyFormat implements TextWatcher {
public void onTextChanged(CharSequence arg0, int start, int arg2,int arg3) {}
public void beforeTextChanged(CharSequence arg0, int start,int arg2, int arg3) {}
public void afterTextChanged(Editable arg0) {
int length = arg0.length();
if(length>0){
if(nrOfDecimal(arg0.toString())>2)
arg0.delete(length-1, length);
}
}
private int nrOfDecimal(String nr){
int len = nr.length();
int pos = len;
for(int i=0 ; i<len; i++){
if(nr.charAt(i)=='.'){
pos=i+1;
break;
}
}
return len-pos;
}
}
答案 23 :(得分：1)
一个很晚的回应： 我们可以这样做：
etv.addTextChangedListener(new TextWatcher() {
@Override
public void beforeTextChanged(CharSequence s, int start, int count, int after) {
}
@Override
public void onTextChanged(CharSequence s, int start, int before, int count) {
if (s.toString().length() > 3 && s.toString().contains(".")) {
if (s.toString().length() - s.toString().indexOf(".") > 3) {
etv.setText(s.toString().substring(0, s.length() - 1));
etv.setSelection(edtSendMoney.getText().length());
}
}
}
@Override
public void afterTextChanged(Editable arg0) {
}
}
答案 24 :(得分：0)
我没有正则表达式的简单解决方案
int start=Edit1.getSelectionStart();
String sp=Edit1.getText().toString();
sp=sp.replace(",",".");
Double d=Double.valueOf(sp);
String s=String.format("%.2f",d );
if(!Edit1.getText().toString().equals(s))
Edit1.setText(s);
if(start>Edit1.getText().length())start--;
Edit1.setSelection(start);
放入onTextChange。删除逗号后，零会翻倍，因为数字变为整数。
答案 25 :(得分：0)
在Android Kotlin中创建一个名为DecimalDigitsInputFilter的新类
class DecimalDigitsInputFilter(digitsBeforeZero: Int, digitsAfterZero: Int) : InputFilter {
lateinit var mPattern: Pattern
init {
mPattern =
Pattern.compile("[0-9]{0," + (digitsBeforeZero) + "}+((\\.[0-9]{0," + (digitsAfterZero) + "})?)||(\\.)?")
}
override fun filter(
source: CharSequence?,
start: Int,
end: Int,
dest: Spanned?,
dstart: Int,
dend: Int
): CharSequence? {
val matcher: Matcher = mPattern.matcher(dest?.subSequence(0, dstart).toString() + source?.subSequence(start, end).toString() + dest?.subSequence(dend, dest?.length!!).toString())
if (!matcher.matches())
return ""
else
return null
}
通过以下行调用该课程
et_buy_amount.filters = (arrayOf<InputFilter>(DecimalDigitsInputFilter(8,2)))
对于相同的答案太多了，但可以让您在小数点前输入8位数字，在小数点后输入2位数字
其他答案仅接受8位数字
答案 26 :(得分：0)
就像其他人说的那样，我在项目中添加了此类，并将过滤器设置为EditText Simpler解决方案，而无需使用正则表达式：
void main(){
struct phone{
char name[100];
char num[10];
};
phone book[100];
for(int i = 0; i<100; i++){
cin>>book[i].name;
cin>>book[i].num;
}
}
}
应用过滤器：
public class DecimalDigitsInputFilter implements InputFilter {
int digitsBeforeZero =0;
int digitsAfterZero=0;
public DecimalDigitsInputFilter(int digitsBeforeZero,int digitsAfterZero) {
this.digitsBeforeZero=digitsBeforeZero;
this.digitsAfterZero=digitsAfterZero;
}
@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
if(dest!=null && dest.toString().trim().length()<(digitsBeforeZero+digitsAfterZero)){
String value=dest.toString().trim();
if(value.contains(".") && (value.substring(value.indexOf(".")).length()<(digitsAfterZero+1))){
return ((value.indexOf(".")+1+digitsAfterZero)>dstart)?null:"";
}else if(value.contains(".") && (value.indexOf(".")<dstart)){
return "";
}else if(source!=null && source.equals(".")&& ((value.length()-dstart)>=(digitsAfterZero+1))){
return "";
}
}else{
return "";
}
return null;
}
答案 27 :(得分：0)
InputFilter的这种实现解决了这个问题。
import android.text.SpannableStringBuilder;
import android.text.Spanned;
import android.text.method.DigitsKeyListener;
public class MoneyValueFilter extends DigitsKeyListener {
public MoneyValueFilter() {
super(false, true);
}
private int digits = 2;
public void setDigits(int d) {
digits = d;
}
@Override
public CharSequence filter(CharSequence source, int start, int end,
Spanned dest, int dstart, int dend) {
CharSequence out = super.filter(source, start, end, dest, dstart, dend);
// if changed, replace the source
if (out != null) {
source = out;
start = 0;
end = out.length();
}
int len = end - start;
// if deleting, source is empty
// and deleting can't break anything
if (len == 0) {
return source;
}
int dlen = dest.length();
// Find the position of the decimal .
for (int i = 0; i < dstart; i++) {
if (dest.charAt(i) == '.') {
// being here means, that a number has
// been inserted after the dot
// check if the amount of digits is right
return (dlen-(i+1) + len > digits) ?
"" :
new SpannableStringBuilder(source, start, end);
}
}
for (int i = start; i < end; ++i) {
if (source.charAt(i) == '.') {
// being here means, dot has been inserted
// check if the amount of digits is right
if ((dlen-dend) + (end-(i + 1)) > digits)
return "";
else
break; // return new SpannableStringBuilder(source, start, end);
}
}
// if the dot is after the inserted part,
// nothing can break
return new SpannableStringBuilder(source, start, end);
}
}
要使用：
editCoin.setFilters(new InputFilter[] {new MoneyValueFilter(2)});
答案 28 :(得分：0)
@Override
public void beforeTextChanged(CharSequence s, int start, int count, int after) {
String numero = total.getText().toString();
int dec = numero.indexOf(".");
int longitud = numero.length();
if (dec+3 == longitud && dec != -1) { //3 number decimal + 1
log.i("ento","si");
numero = numero.substring(0,dec+3);
if (contador == 0) {
contador = 1;
total.setText(numero);
total.setSelection(numero.length());
} else {
contador = 0;
}
}
}
答案 29 :(得分：0)
实现这一目标的最简单方法是：
et.addTextChangedListener(new TextWatcher() {
public void onTextChanged(CharSequence arg0, int arg1, int arg2, int arg3) {
String text = arg0.toString();
if (text.contains(".") && text.substring(text.indexOf(".") + 1).length() > 2) {
et.setText(text.substring(0, text.length() - 1));
et.setSelection(et.getText().length());
}
}
public void beforeTextChanged(CharSequence arg0, int arg1, int arg2, int arg3) {
}
public void afterTextChanged(Editable arg0) {
}
});
答案 30 :(得分：0)
这是建立在pinhassi的答案上 - 我遇到的问题是，一旦达到小数限制，就无法在小数点前添加值。要解决这个问题，我们需要在进行模式匹配之前构造最终的字符串。
import java.util.regex.Matcher;
import java.util.regex.Pattern;
import android.text.InputFilter;
import android.text.Spanned;
public class DecimalLimiter implements InputFilter
{
Pattern mPattern;
public DecimalLimiter(int digitsBeforeZero,int digitsAfterZero)
{
mPattern=Pattern.compile("[0-9]{0," + (digitsBeforeZero) + "}+((\\.[0-9]{0," + (digitsAfterZero) + "})?)||(\\.)?");
}
@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend)
{
StringBuilder sb = new StringBuilder(dest);
sb.insert(dstart, source, start, end);
Matcher matcher = mPattern.matcher(sb.toString());
if(!matcher.matches())
return "";
return null;
}
}
答案 31 :(得分：0)
以下TextWatcher只允许 n 小数点后的位数。
<强> TextWatcher
private static boolean flag;
public static TextWatcher getTextWatcherAllowAfterDeci(final int allowAfterDecimal){
TextWatcher watcher = new TextWatcher() {
@Override
public void onTextChanged(CharSequence s, int start, int before, int count) {
// TODO Auto-generated method stub
}
@Override
public void beforeTextChanged(CharSequence s, int start, int count,
int after) {
// TODO Auto-generated method stub
}
@Override
public void afterTextChanged(Editable s) {
// TODO Auto-generated method stub
String str = s.toString();
int index = str.indexOf ( "." );
if(index>=0){
if((index+1)<str.length()){
String numberD = str.substring(index+1);
if (numberD.length()!=allowAfterDecimal) {
flag=true;
}else{
flag=false;
}
}else{
flag = false;
}
}else{
flag=false;
}
if(flag)
s.delete(s.length() - 1,
s.length());
}
};
return watcher;
}
如何使用
yourEditText.addTextChangedListener(getTextWatcherAllowAfterDeci(1));
答案 32 :(得分：0)
这对我来说很好。即使在焦点更改和检索后，它也允许输入值。例如：123.00，12.12，0.01等。
1。Integer.parseInt(getString(R.string.valuelength))
指定从digits.Values文件访问的输入string.xml的长度。它很容易更改值。
2。Integer.parseInt(getString(R.string.valuedecimal))，这是小数位数最大限制。
private InputFilter[] valDecimalPlaces;
private ArrayList<EditText> edittextArray;
valDecimalPlaces = new InputFilter[] { new DecimalDigitsInputFilterNew(
Integer.parseInt(getString(R.string.valuelength)),
Integer.parseInt(getString(R.string.valuedecimal)))
};
允许执行操作的EditText值数组。
for (EditText etDecimalPlace : edittextArray) {
etDecimalPlace.setFilters(valDecimalPlaces);
我刚使用了包含多个edittext的值数组
下一个DecimalDigitsInputFilterNew.class文件。
import android.text.InputFilter;
import android.text.Spanned;
public class DecimalDigitsInputFilterNew implements InputFilter {
private final int decimalDigits;
private final int before;
public DecimalDigitsInputFilterNew(int before ,int decimalDigits) {
this.decimalDigits = decimalDigits;
this.before = before;
}
@Override
public CharSequence filter(CharSequence source, int start, int end,
Spanned dest, int dstart, int dend) {
StringBuilder builder = new StringBuilder(dest);
builder.replace(dstart, dend, source
.subSequence(start, end).toString());
if (!builder.toString().matches("(([0-9]{1})([0-9]{0,"+(before-1)+"})?)?(\\.[0-9]{0,"+decimalDigits+"})?")) {
if(source.length()==0)
return dest.subSequence(dstart, dend);
return "";
}
return null;
}
}
答案 33 :(得分：0)
et = (EditText) vw.findViewById(R.id.tx_edittext);
et.setFilters(new InputFilter[] {
new DigitsKeyListener(Boolean.FALSE, Boolean.TRUE) {
int beforeDecimal = 5, afterDecimal = 2;
@Override
public CharSequence filter(CharSequence source, int start, int end,
Spanned dest, int dstart, int dend) {
String temp = et.getText() + source.toString();
if (temp.equals(".")) {
return "0.";
}
else if (temp.toString().indexOf(".") == -1) {
// no decimal point placed yet
if (temp.length() > beforeDecimal) {
return "";
}
} else {
temp = temp.substring(temp.indexOf(".") + 1);
if (temp.length() > afterDecimal) {
return "";
}
}
return super.filter(source, start, end, dest, dstart, dend);
}
}
});
答案 34 :(得分：0)
这是我的解决方案：
yourEditText.addTextChangedListener(new TextWatcher() {
@Override
public void onTextChanged(CharSequence s, int start, int before, int count) {
NumberFormat formatter = new DecimalFormat("#.##");
double doubleVal = Double.parseDouble(s.toString());
yourEditText.setText(formatter.format(doubleVal));
}
@Override
public void beforeTextChanged(CharSequence s, int start, int count,int after) {}
@Override
public void afterTextChanged(Editable s) {}
});
如果用户在小数点后输入一个数字超过两位的数字，它将自动更正。
我希望我有所帮助！
答案 35 :(得分：-1)
对于Kotlin
val inputFilter = arrayOf<InputFilter>(DecimalDigitsInputFilter(5,2))
et_total_value.setFilters(inputFilter)
答案 36 :(得分：-3)
这是将小数点后的位数限制为2的最简单的解决方案：
myeditText2 = (EditText) findViewById(R.id.editText2);
myeditText2.setInputType(3);