我试图实现限制输入到阿尔法字符只有[A-ZA-Z]一个EditText。
我开始从输入过滤方法这篇文章 。 当我键入“A％”的文本，然后消失，如果我打退格文是“A”。 我试过在过滤器功能的其他变化就像使用正则表达式只匹配[A-ZA-Z]，有时看到这样重复字符的疯狂行为，我将输入“A”，然后“B”，并获得“AAB”，然后“C”型，并获得“aabaabc”然后按退格并获得“aabaabcaabaabc”！
下面是我与迄今与我尝试了不同的方法使用的代码。
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 } );
我完全难倒这一个所以这里的任何帮助，将不胜感激。
编辑：好的，我已经做了很多与输入过滤实验，并得出了一些结论，虽然没有解决的问题。 请参阅我下面的代码中的注释。 我现在要去尝试伊姆兰林蛙的解决方案。
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 } );
使用下面的代码：
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
}
});
如果你想有效的文字留在EditText上：
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,"");
}
}
}
});
我们也有类似的问题，我相信一个解决方案[0]会为你正常工作。 我们的要求是实现一个剥离富文本输入一个EditText。 例如，如果用户复制粗体来他们的剪贴板并粘贴入的EditText，在EditText上应该删除大胆强调造型和仅保留纯文本。
该解决方案类看起来是这样的：
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) {
// ...
}
}
}
我们最初的实施stripRichText（）很简单：
// -- BROKEN. DO NOT USE --
String plainText = str.subSequence(start, end).toString();
return plainText;
Java的基础String类不保留任何样式的信息，以便为CharSequence接口转换为具体的字符串拷贝只纯文本。
我们没有意识到的是，部分Android软键盘添加和依靠错别字和其他东西临时组成的提示。 该问题通过去除提示以及以意想不到的方式（通常加倍整个的EditText字段的输入）重复字符表现。 对于InputFilter.filter文档[1]（）进行通信的要求是这样的：
* 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).
我相信，妥善解决是保存临时跨度：
/** 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;
}
[0]可以在这里找到实际实现： https://git.wikimedia.org/blob/apps%2Fandroid%2Fwikipedia/e9ddd8854ff15cde791a2e6fb7754a5450d6f7cf/app%2Fsrc%2Fmain%2Fjava%2Forg%2Fwikipedia%2Frichtext%2FRichTextUtil.java
[1] https://android.googlesource.com/platform/frameworks/base/+/029942f77d05ed3d20256403652b220c83dad6e1/core/java/android/text/InputFilter.java#37
宾果，我发现这个问题！
当我使用Android：在EditText上cursorVisible =“假”的开始和DSTART参数不正确匹配。
启动参数仍然是我始终为0，但DSTART参数也总是0，因此只要我使用.replaceAll无效（）。 这是相反的是这个帖子是这样说的我不太明白为什么，但至少我可以建立一些作品吧！
我只是想给我的解决方案添加到问题（如迟到，因为它是）。 我发现，如果添加
yourEditText.setInputType(InputType.TYPE_TEXT_FLAG_NO_SUGGESTIONS);
然后退格问题停止
修正了重复的文字，工作在所有的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;
}
};
}
现在你可以使用这个文件管理器，如：
//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)});
