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java 判断字符串中是否包含emoj表情及过滤,完美解决

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代码如下:

package com.luo.dtqjh.utils;

import org.apache.commons.lang.StringUtils;

public class EmojiFilter {
public static boolean containsEmoji(String source) {
int len = source.length();
boolean isEmoji = false;
for (int i = 0; i < len; i++) {
char hs = source.charAt(i);
if (0xd800 <= hs && hs <= 0xdbff) {
if (source.length() > 1) {
char ls = source.charAt(i + 1);
int uc = ((hs - 0xd800) * 0x400) + (ls - 0xdc00) + 0x10000;
if (0x1d000 <= uc && uc <= 0x1f77f) {
return true;
}
}
} else {
// non surrogate
if (0x2100 <= hs && hs <= 0x27ff && hs != 0x263b) {
return true;
} else if (0x2B05 <= hs && hs <= 0x2b07) {
return true;
} else if (0x2934 <= hs && hs <= 0x2935) {
return true;
} else if (0x3297 <= hs && hs <= 0x3299) {
return true;
} else if (hs == 0xa9 || hs == 0xae || hs == 0x303d
|| hs == 0x3030 || hs == 0x2b55 || hs == 0x2b1c
|| hs == 0x2b1b || hs == 0x2b50 || hs == 0x231a) {
return true;
}
if (!isEmoji && source.length() > 1 && i < source.length() - 1) {
char ls = source.charAt(i + 1);
if (ls == 0x20e3) {
return true;
}
}
}
}
return isEmoji;
}

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));
}

/**
* 过滤emoji 或者 其他非文字类型的字符
*
* @param source
* @return
*/
public static String filterEmoji(String source) {
if (StringUtils.isBlank(source)) {
return source;
}
StringBuilder buf = null;
int len = source.length();
for (int i = 0; i < len; i++) {
char codePoint = source.charAt(i);
if (isEmojiCharacter(codePoint)) {
if (buf == null) {
buf = new StringBuilder(source.length());
}
buf.append(codePoint);
}
}
if (buf == null) {
return source;
} else {
if (buf.length() == len) {
buf = null;
return source;
} else {
return buf.toString();
}
}
}
}


测试代码:

public static void main(String[] args) {
String string = "��都嗨��、齐静��给你��";
System.out.println(EmojiFilter.containsEmoji(string));
System.out.println(EmojiFilter.filterEmoji(string));
}


2019-09-10 14:23:21     阅读(1230)

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