ajax乱码解决

public static String unescape(String src) {
		StringBuffer tmp = new StringBuffer();
		tmp.ensureCapacity(src.length());
		int lastPos = 0, pos = 0;
		char ch;
		while (lastPos < src.length()) {
			pos = src.indexOf("%", lastPos);
			if (pos == lastPos) {
				if (src.charAt(pos + 1) == 'u') {
					ch = (char) Integer.parseInt(src
							.substring(pos + 2, pos + 6), 16);
					tmp.append(ch);
					lastPos = pos + 6;
				} else {
					ch = (char) Integer.parseInt(src
							.substring(pos + 1, pos + 3), 16);
					tmp.append(ch);
					lastPos = pos + 3;
				}
			} else {
				if (pos == -1) {
					tmp.append(src.substring(lastPos));
					lastPos = src.length();
				} else {
					tmp.append(src.substring(lastPos, pos));
					lastPos = pos;
				}
			}
		}
		return tmp.toString();
	}

 在发送ajax请求之前,对参数escape()两次( 例:name=escape(escape("张三")); ),然后再发送请求到服务器,服务器接收参数后,对参数unescape(),就可以得到正确的参数( 例:String name=unescape(request.getParameter("name")); ).

 

附java版的unescape实现