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@ywjno
Created January 27, 2015 07:57
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EmojiFilter.java
import org.apache.commons.lang.StringUtils;
public class EmojiFilter {
/**
* 检测是否有emoji字符
* @param source
* @return 一旦含有就抛出
*/
public boolean containsEmoji(String source) {
if (StringUtils.isBlank(source)) {
return false;
}
int len = source.length();
for (int i = 0; i < len; i++) {
char codePoint = source.charAt(i);
if (isEmojiCharacter(codePoint)) {
//do nothing,判断到了这里表明,确认有表情字符
return true;
}
}
return false;
}
private 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 String filterEmoji(String source) {
if (!containsEmoji(source)) {
return source;//如果不包含,直接返回
}
//到这里铁定包含
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);
} else {
}
}
if (buf == null) {
return source;//如果没有找到 emoji表情,则返回源字符串
} else {
if (buf.length() == len) {//这里的意义在于尽可能少的toString,因为会重新生成字符串
buf = null;
return source;
} else {
return buf.toString();
}
}
}
}
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