Question

Ajax提交参数时html标签提交失败如何处理

Answer 1

这种处理方法可行，把html符号过滤替换，后台接收时再替换回去。
比如需要提交的变量message，替换如下：

message=message.replace(/</g,"@左尖括号@");
message=message.replace(/>/g,"@右尖括号@");
message=message.replace(/\//g,"@左斜杠@");
message=message.replace(/=/g,"@等号@");
message=message.replace(/&/g,"@and@");
message=message.replace(/\"/g,"@双引号@");
message=message.replace(/ /g,"@空格@");
替换完毕后发送，后台接收数据时候又替换回去。测试可用。

Answer 2

方法
function base64decode(str) {
var c1, c2, c3, c4;
var i, len, out;
len = str.length;
i = 0;
out = “”;
while(i < len) {
do {
c1 = base64DecodeChars[str.charCodeAt(i++) & 0xff];
} while(i < len && c1 == -1);
if(c1 == -1)
break;
do {
c2 = base64DecodeChars[str.charCodeAt(i++) & 0xff];
} while(i < len && c2 == -1);
if(c2 == -1)
break;
out += String.fromCharCode((c1 << 2) | ((c2 & 0x30) >> 4));
do {
c3 = str.charCodeAt(i++) & 0xff;
if(c3 == 61)
return out;
c3 = base64DecodeChars[c3];
} while(i < len && c3 == -1);
if(c3 == -1)
break;
out += String.fromCharCode(((c2 & 0XF) << 4) | ((c3 & 0x3C) >> 2));
do {
c4 = str.charCodeAt(i++) & 0xff;
if(c4 == 61)
return out;
c4 = base64DecodeChars[c4];
} while(i < len && c4 == -1);
if(c4 == -1)
break;
out += String.fromCharCode(((c3 & 0x03) << 6) | c4);
}
return out;
}
function utf16to8(str) {
var out, i, len, c;
out = “”;
len = str.length;
for(i = 0; i < len; i++) {
c = str.charCodeAt(i);
if ((c >= 0x0001) && (c <= 0x007F)) {
out += str.charAt(i);
} else if (c > 0x07FF) {
out += String.fromCharCode(0xE0 | ((c >> 12) & 0x0F));
out += String.fromCharCode(0x80 | ((c >> 6) & 0x3F));
out += String.fromCharCode(0x80 | ((c >> 0) & 0x3F));
} else {
out += String.fromCharCode(0xC0 | ((c >> 6) & 0x1F));
out += String.fromCharCode(0x80 | ((c >> 0) & 0x3F));
}
}
return out;
}
function utf8to16(str) {
var out, i, len, c;
var char2, char3;
out = “”;
len = str.length;
i = 0;
while(i < len) {
c = str.charCodeAt(i++);
switch(c >> 4)
{
case 0: case 1: case 2: case 3: case 4: case 5: case 6: case 7:
// 0xxxxxxx
out += str.charAt(i-1);
break;
case 12: case 13:
// 110x xxxx 10xx xxxx
char2 = str.charCodeAt(i++);
out += String.fromCharCode(((c & 0x1F) << 6) | (char2 & 0x3F));
break;
case 14:
// 1110 xxxx 10xx xxxx 10xx xxxx
char2 = str.charCodeAt(i++);
char3 = str.charCodeAt(i++);
out += String.fromCharCode(((c & 0x0F) << 12) |
((char2 & 0x3F) << 6) |
((char3 & 0x3F) << 0));
break;
}
}
return out;