我正在尝试实现一个EditText，它只将输入限制为alpha字符[A-Za-z]。
我从this post开始使用InputFilter方法。当我输入“a％”时，文本消失，如果我按退格键，则文本为“a”。我已经尝试了过滤器功能的其他变体，比如使用正则表达式只匹配[A-Za-z]，有时会看到像重复字符这样的疯狂行为，我会输入“a”然后“b”并获得“aab”然后输入“c”并获取“aabaabc”然后点击退格并获得“aabaabcaabaabc”！
这是我到目前为止使用的代码，我尝试过不同的方法。
EditText input = (EditText)findViewById( R.id.inputText );
InputFilter filter = new InputFilter() {
@Override
public CharSequence filter( CharSequence source, int start, int end, Spanned dest, int dstart, int dend ) {
//String data = source.toString();
//String ret = null;
/*
boolean isValid = data.matches( "[A-Za-z]" );
if( isValid ) {
ret = null;
}
else {
ret = data.replaceAll( "[@#$%^&*]", "" );
}
*/
/*
dest = new SpannableStringBuilder();
ret = data.replaceAll( "[@#$%^&*]", "" );
return ret;
*/
for( int i = start; i < end; i++ ) {
if( !Character.isLetter( source.charAt( i ) ) ) {
return "";
}
}
return null;
}
};
input.setFilters( new InputFilter[]{ filter } );
我对这一点感到非常难过，所以非常感谢这里的任何帮助。
修改 好吧，我已经做了大量的InputFilter实验并得出了一些结论，尽管没有解决问题的办法。请参阅下面的代码中的注释。我现在要尝试Imran Rana的解决方案。
EditText input = (EditText)findViewById( R.id.inputText );
InputFilter filter = new InputFilter() {
// It is not clear what this function should return!
// Docs say return null to allow the new char(s) and return "" to disallow
// but the behavior when returning "" is inconsistent.
//
// The source parameter is a SpannableStringBuilder if 1 char is entered but it
// equals the whole string from the EditText.
// If more than one char is entered (as is the case with some keyboards that auto insert
// a space after certain chars) then the source param is a CharSequence and equals only
// the new chars.
@Override
public CharSequence filter( CharSequence source, int start, int end, Spanned dest, int dstart, int dend ) {
String data = source.toString().substring( start, end );
String retData = null;
boolean isValid = data.matches( "[A-Za-z]+" );
if( !isValid ) {
if( source instanceof SpannableStringBuilder ) {
// This works until the next char is evaluated then you get repeats
// (Enter "a" then "^" gives "a". Then enter "b" gives "aab")
retData = data.replaceAll( "[@#$%^&*']", "" );
// If I instead always returns an empty string here then the EditText is blanked.
// (Enter "a" then "^" gives "")
//retData = "";
}
else { // source is instanceof CharSequence
// We only get here if more than 1 char was entered (like "& ").
// And again, this works until the next char is evaluated then you get repeats
// (Enter "a" then "& " gives "a". Then enter "b" gives "aab")
retData = "";
}
}
return retData;
}
};
input.setFilters( new InputFilter[]{ filter } );
答案 0 :(得分：4)
EditText input = (EditText) findViewById(R.id.inputText);
input.addTextChangedListener(new TextWatcher() {
public void onTextChanged(CharSequence s, int start, int before, int count) {
// TODO Auto-generated method stub
for( int i = start;i<s.toString().length(); i++ ) {
if( !Character.isLetter(s.charAt( i ) ) ) {
input.setText("");
}
}
}
public void beforeTextChanged(CharSequence s, int start, int count,
int after) {
// TODO Auto-generated method stub
}
public void afterTextChanged(Editable s) {
// TODO Auto-generated method stub
}
});
input.addTextChangedListener(new TextWatcher() {
public void onTextChanged(CharSequence s, int start, int before, int count) {
// TODO Auto-generated method stub
}
public void beforeTextChanged(CharSequence s, int start, int count,
int after) {
// TODO Auto-generated method stub
}
public void afterTextChanged(Editable s) {
// TODO Auto-generated method stub
for( int i = 0;i<s.toString().length(); i++ ) {
if( !Character.isLetter(s.charAt( i ) ) ) {
s.replace(i, i+1,"");
}
}
}
});
答案 1 :(得分：2)
我们遇到了类似的问题，我相信一个适合您的解决方案[0]。我们的要求是实现一个剥离富文本输入的EditText。例如，如果用户将粗体文本复制到剪贴板并将其粘贴到EditText中，则EditText应删除粗体强调样式并仅保留纯文本。
解决方案类看起来像这样：
public class PlainEditText extends EditText {
public PlainEditText(Context context, AttributeSet attrs, int defStyleAttr) {
super(context, attrs, defStyleAttr);
addFilter(this, new PlainTextInputFilter());
}
private void addFilter(TextView textView, InputFilter filter) {
InputFilter[] filters = textView.getFilters();
InputFilter[] newFilters = Arrays.copyOf(filters, filters.length + 1);
newFilters[filters.length] = filter;
textView.setFilters(newFilters);
}
private static class PlainTextInputFilter implements InputFilter {
@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest,
int dstart, int dend) {
return stripRichText(source, start, end);
}
private CharSequence stripRichText(CharSequence str, int start, int end) {
// ...
}
}
}
我们对stripRichText（）的原始实现很简单：
// -- BROKEN. DO NOT USE --
String plainText = str.subSequence(start, end).toString();
return plainText;
Java基类String类不保留任何样式信息，因此将CharSequence接口转换为具体的String只复制纯文本。
我们没有意识到的是，一些Android软键盘会添加并依赖于拼写错误和其他内容的临时构图提示。问题通过删除提示以及以意外方式重复字符（通常将整个EditText字段的输入加倍）来体现。 InputFilter.filter（）的文档[1]以这种方式传达需求：
* Note: If <var>source</var> is an instance of {@link Spanned} or
* {@link Spannable}, the span objects in the <var>source</var> should be
* copied into the filtered result (i.e. the non-null return value).
我认为正确的解决办法是保持临时跨度：
/** Strips all rich text except spans used to provide compositional hints. */
private CharSequence stripRichText(CharSequence str, int start, int end) {
String plainText = str.subSequence(start, end).toString();
SpannableString ret = new SpannableString(plainText);
if (str instanceof Spanned) {
List<Object> keyboardHintSpans = getComposingSpans((Spanned) str, start, end);
copySpans((Spanned) str, ret, keyboardHintSpans);
}
return ret;
}
/**
* @return Temporary spans, often applied by the keyboard to provide hints such as typos.
*
* @see {@link android.view.inputmethod.BaseInputConnection#removeComposingSpans}
* @see {@link android.inputmethod.latin.inputlogic.InputLogic#setComposingTextInternalWithBackgroundColor}
*/
@NonNull private List<Object> getComposingSpans(@NonNull Spanned spanned,
int start,
int end) {
// TODO: replace with Apache CollectionUtils.filter().
List<Object> ret = new ArrayList<>();
for (Object span : getSpans(spanned, start, end)) {
if (isComposingSpan(spanned, span)) {
ret.add(span);
}
}
return ret;
}
private Object[] getSpans(@NonNull Spanned spanned, int start, int end) {
Class<Object> anyType = Object.class;
return spanned.getSpans(start, end, anyType);
}
private boolean isComposingSpan(@NonNull Spanned spanned, Object span) {
return isFlaggedSpan(spanned, span, Spanned.SPAN_COMPOSING);
}
private boolean isFlaggedSpan(@NonNull Spanned spanned, Object span, int flags) {
return (spanned.getSpanFlags(span) & flags) == flags;
}
答案 2 :(得分：1)
当我在EditText上使用android：cursorVisible =“false”时，start和dstart参数无法正确匹配。
对于我来说start参数仍然总是0，但是dstart参数也总是0，所以只要我使用.replaceAll（）就可以运行。这与this post所说的相反，所以我不太明白为什么，但至少我可以建立一些现在有用的东西！
答案 3 :(得分：0)
我只想添加我的解决方案（尽可能晚）。我发现如果你添加
yourEditText.setInputType(InputType.TYPE_TEXT_FLAG_NO_SUGGESTIONS);
然后退格问题停止
答案 4 :(得分：0)
修复重复文字，适用于所有Android版本：
public static InputFilter getOnlyCharactersFilter() {
return getCustomInputFilter(true, false, false);
}
public static InputFilter getCharactersAndDigitsFilter() {
return getCustomInputFilter(true, true, false);
}
public static InputFilter getCustomInputFilter(final boolean allowCharacters, final boolean allowDigits, final boolean allowSpaceChar) {
return new InputFilter() {
@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
boolean keepOriginal = true;
StringBuilder sb = new StringBuilder(end - start);
for (int i = start; i < end; i++) {
char c = source.charAt(i);
if (isCharAllowed(c)) {
sb.append(c);
} else {
keepOriginal = false;
}
}
if (keepOriginal) {
return null;
} else {
if (source instanceof Spanned) {
SpannableString sp = new SpannableString(sb);
TextUtils.copySpansFrom((Spanned) source, start, sb.length(), null, sp, 0);
return sp;
} else {
return sb;
}
}
}
private boolean isCharAllowed(char c) {
if (Character.isLetter(c) && allowCharacters) {
return true;
}
if (Character.isDigit(c) && allowDigits) {
return true;
}
if (Character.isSpaceChar(c) && allowSpaceChar) {
return true;
}
return false;
}
};
}
现在您可以使用此文件管理器：
//Accept Characters Only
edit_text.setFilters(new InputFilter[]{getOnlyCharactersFilter()});
//Accept Digits and Characters
edit_text.setFilters(new InputFilter[]{getCharactersAndDigitsFilter()});
//Accept Digits and Characters and SpaceBar
edit_text.setFilters(new InputFilter[]{getCustomInputFilter(true,true,true)});