我使用下面的InputFilter过滤掉无效(十进制值大于127的ASCII),并且EditText在输入无效字符时显示文本两次。我希望EditText能够显示有效的字符,下面是发生什么的例子。创建扩展的ASCII字符InputFilter
- 用户进入XYZ€中的EditText组件的屏幕提示“无效的非ASCII字符”
-EditText组件显示在屏幕上的xyz这是我所期待的
上出现
-toast消息 - 用户输入有效字符,甲因此屏幕显示XYZA
-InputFilter运行并返回XYZA但XYZXYZA出现在的EditText部件whic h不正确。它复制了XYZ
有关为什么在处理无效字符后复制输入文本的任何想法?
屏幕:
<EditText android:id="@+id/editText"
android:layout_width="fill_parent" android:layout_height="120dp"
android:layout_marginTop="10dp" android:layout_marginLeft="10dp"
android:layout_marginRight="10dp" android:maxLength="45"
android:focusable="true" android:inputType="text"
android:cursorVisible="true" android:imeOptions="actionDone"
/>
活动:
public class EditTextActivity extends Activity {
@Override
protected void onCreate(Bundle savedInstanceState) {
EditText eText = (EditText) findViewById(R.id.editText);
setAsciiTextFilter();
}
private void setAsciiTextFilter() {
InputFilter filter = new InputFilter() {
int asciiNo;
@Override
public CharSequence filter(CharSequence source, int start, int end,
Spanned dest, int dstart, int dend) {
for (int i = start; i < end; i++) {
asciiNo = source.charAt(i);
if(asciiNo > 127) {
toast = mUtility.showToast("Invalid non-Ascii Character", Toast.LENGTH_SHORT);
//Replace the invalid ascii character with empty String
return source.toString().replace(source.charAt(i)+"", "");
}
}
return null;
}
};
eText.setFilters(new InputFilter[]{filter});
}
}
新尝试:
InputFilter filter = new InputFilter() {
@Override
public CharSequence filter(CharSequence source, int start, int end,
Spanned dest, int dstart, int dend) {
if (source instanceof SpannableStringBuilder) {
SpannableStringBuilder sourceAsSpannableBuilder = (SpannableStringBuilder)source;
for (int i = end - 1; i >= start; i--) {
char currentChar = source.charAt(i);
int ascii = currentChar;
if (ascii > 127) {
sourceAsSpannableBuilder.delete(i, i+1);
toast = mUtility.showToast("Invalid non-Ascii Character", Toast.LENGTH_SHORT);
}
}
return source;
} else {
StringBuilder filteredStringBuilder = new StringBuilder();
for (int i = 0; i < end; i++) {
char currentChar = source.charAt(i);
int ascii = currentChar;
if (ascii <= 127) {
filteredStringBuilder.append(currentChar);
}
}
return filteredStringBuilder.toString();
}
}
};
感谢您的回答。我使用了你的建议并更新了我的代码(在新的尝试:部分)。由于sourceAsSpannableBuilder.delete(i,i + 1);一旦输入了无效字符,即使有效字符,toast也会继续显示。每次都会不断移除无效字符。尽管EditText不显示它,但SourceAsSpannableBuilder似乎仍保留无效字符。只有在输入无效字符时才显示敬酒的任何方法?感谢所有帮助... – c12
无视上面的问题,如果(结束==我+ 1){/ /显示烤面包}作品 – c12