Question

如何在Android中的EditText中只输入字母？

Answer 1

使用EditText标记添加此行。

android:digits="abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"

您的EditText标记应如下所示：

<EditText
        android:id="@+id/editText1"
        android:digits="abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
        android:layout_width="fill_parent"
        android:layout_height="wrap_content" />

Answer 2

edittext.setFilters(new InputFilter[] {
    new InputFilter() {
        public CharSequence filter(CharSequence src, int start,
                int end, Spanned dst, int dstart, int dend) {
            if(src.equals("")){ // for backspace
                return src;
            }
            if(src.toString().matches("[a-zA-Z ]+")){
                return src;
            }
            return "";
        }
    }
});

请尽快测试！

Answer 3

EditText state = (EditText) findViewById(R.id.txtState);


                Pattern ps = Pattern.compile("^[a-zA-Z ]+$");
                Matcher ms = ps.matcher(state.getText().toString());
                boolean bs = ms.matches();
                if (bs == false) {
                    if (ErrorMessage.contains("invalid"))
                        ErrorMessage = ErrorMessage + "state,";
                    else
                        ErrorMessage = ErrorMessage + "invalid state,";

                }

Answer 4

对于那些希望他们的editText只接受字母和空格（无数字或任何特殊字符）的人，可以使用此InputFilter。在这里，我创建了一个名为getEditTextFilter()的方法，并在其中编写了InputFilter。

public static InputFilter getEditTextFilter() {
        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)) // put your condition here
                        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) {
                Pattern ps = Pattern.compile("^[a-zA-Z ]+$");
                Matcher ms = ps.matcher(String.valueOf(c));
                return ms.matches();
            }
        };
    }

找到后将此inputFilter附加到editText，如下所示：

mEditText.setFilters(new InputFilter[]{getEditTextFilter()});

最初的功劳归于@UMAR，他提出了使用正则表达式和@KamilSeweryn进行验证的想法

Answer 5

通过Xml，您可以轻松地按照xml（editText）中的代码进行操作...

android:digits="abcdefghijklmnopqrstuvwxyz"

只会接受字符...

Answer 6

放置代码edittext xml文件，

   android:digits="abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"

Answer 7

仅允许EditText中的字母表安卓：

InputFilter letterFilter = new InputFilter() {
        public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
            String filtered = "";
            for (int i = start; i < end; i++) {
                char character = source.charAt(i);
                if (!Character.isWhitespace(character)&&Character.isLetter(character)) {
                    filtered += character;
                }
            }

            return filtered;
        }

    };
editText.setFilters(new InputFilter[]{letterFilter});

Answer 8

对于空格，您可以在数字中添加单个空格。如果您需要任何特殊字符，如点，逗号也可以添加到此列表

android：digits =“abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ”

Answer 9

试试这个

<EditText
  android:id="@+id/EditText1"
  android:text=""
  android:inputType="text|textNoSuggestions"
  android:textSize="18sp"
  android:layout_width="80dp"
  android:layout_height="43dp">
</EditText>

可以找到其他inputType Here ..

Answer 10

如果有人仍然想要这个，Java regex for support Unicode?是个好人。当你想要只有字母（无论是什么编码 - japaneese，瑞典语）时，它就是一个EditText。之后，您可以使用Matcher和Pattern.compile()

进行检查

Answer 11

试试这个方法

对于 Java：

EditText yourEditText = (EditText) findViewById(R.id.yourEditText);
yourEditText.setFilters(new InputFilter[] {
new InputFilter() {
    @Override
    public CharSequence filter(CharSequence cs, int start,
                int end, Spanned spanned, int dStart, int dEnd) {
        // TODO Auto-generated method stub
        if(cs.equals("")){ // for backspace
             return cs;
        }
        if(cs.toString().matches("[a-zA-Z ]+")){
             return cs;
        }
        return "";
    }
}});

对于 Kotlin：

 val yourEditText = findViewById<View>(android.R.id.yourEditText) as EditText
    val reges = Regex("^[0-9a-zA-Z ]+$")
    //this will allow user to only write letter and white space
    yourEditText.filters = arrayOf<InputFilter>(
        object : InputFilter {
            override fun filter(
                cs: CharSequence, start: Int,
                end: Int, spanned: Spanned?, dStart: Int, dEnd: Int,
            ): CharSequence? {
                if (cs == "") { // for backspace
                    return cs
                }
                return if (cs.toString().matches(reges)) {
                    cs
                } else ""
            }
        }
    )

如何创建EditText仅在android中接受Alphabets？

11 个答案: