转载地址:http://blog.csdn.net/elsdnwn/article/details/45390771
这个方法只能过滤掉一部分表情,但是一些键盘原生的表情还是没有办法过滤。
下面给出一个自定义的过滤表情的editText类:
import android.content.Context; import android.text.Editable; import android.text.Selection; import android.text.Spannable; import android.text.TextWatcher; import android.util.AttributeSet; import android.widget.EditText; import android.widget.Toast; public class ContainsEmojiEditText extends EditText { // 输入表情前的光标位置 private int cursorPos; // 输入表情前EditText中的文本 private String inputAfterText; // 是否重置了EditText的内容 private boolean resetText; private Context mContext; public ContainsEmojiEditText(Context context) { super(context); this.mContext = context; initEditText(); } public ContainsEmojiEditText(Context context, AttributeSet attrs) { super(context, attrs); this.mContext = context; initEditText(); } public ContainsEmojiEditText(Context context, AttributeSet attrs, int defStyleAttr) { super(context, attrs, defStyleAttr); this.mContext = context; initEditText(); } // 初始化edittext 控件 private void initEditText() { addTextChangedListener(new TextWatcher() { @Override public void beforeTextChanged(CharSequence s, int start, int before, int count) { if (!resetText) { cursorPos = getSelectionEnd(); // 这里用s.toString()而不直接用s是因为如果用s, // 那么,inputAfterText和s在内存中指向的是同一个地址,s改变了, // inputAfterText也就改变了,那么表情过滤就失败了 inputAfterText = s.toString(); } } @Override public void onTextChanged(CharSequence s, int start, int before, int count) { if (!resetText) { if (count >= 2) {// 表情符号的字符长度最小为2 CharSequence input = s.subSequence(cursorPos, cursorPos + count); if (containsEmoji(input.toString())) { resetText = true; Toast.makeText(mContext, "不支持输入Emoji表情符号", Toast.LENGTH_SHORT).show(); // 是表情符号就将文本还原为输入表情符号之前的内容 setText(inputAfterText); CharSequence text = getText(); if (text instanceof Spannable) { Spannable spanText = (Spannable) text; Selection.setSelection(spanText, text.length()); } } } } else { resetText = false; } } @Override public void afterTextChanged(Editable editable) { } }); } /** * 检测是否有emoji表情 * * @param source * @return */ public static boolean containsEmoji(String source) { int len = source.length(); for (int i = 0; i < len; i++) { char codePoint = source.charAt(i); if (!isEmojiCharacter(codePoint)) { // 如果不能匹配,则该字符是Emoji表情 return true; } } return false; } /** * 判断是否是Emoji * * @param codePoint * 比较的单个字符 * @return */ private static boolean isEmojiCharacter(char codePoint) { return (codePoint == 0x0) || (codePoint == 0x9) || (codePoint == 0xA) || (codePoint == 0xD) || ((codePoint >= 0x20) && (codePoint <= 0xD7FF)) || ((codePoint >= 0xE000) && (codePoint <= 0xFFFD)) || ((codePoint >= 0x10000) && (codePoint <= 0x10FFFF)); } }isEmojiCharacter()判断是否为emjo表情还是不太全面。希望大家有更好的办法可以分享下。