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  • android非法字符的判定、表情符号的判定

    public class EmojiEditText 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 EmojiEditText(Context context) {
    super(context);
    this.mContext = context;
    initEditText();
    }


    public EmojiEditText(Context context, AttributeSet attrs) {
    super(context, attrs);
    this.mContext = context;
    initEditText();
    }


    public EmojiEditText(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;
    //暂不支持输入Emoji表情符号
    Toast.makeText(mContext, "暂不支持输入表情符号",
    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));
    }


    }

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  • 原文地址:https://www.cnblogs.com/ldq2016/p/8665624.html
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