Question

大家好我正在构建一个存储为varible的URL，以便在单击时可用于选项卡。

URL的参数是一些常量和字符串。所以在一个参数中，用户已经输入了Textarea coulmn

This is
to be
removed 
immediately

等构建URL，浏览器正在抛出'未终止的字符串文字'

以下是“查看来源”

中显示的网址

<script type="text/javascript">var tab3700Action = "/PM/action/showTaskCompletionStatus?preventSTIES=
&deadlinedate=4/22/2005&date=4/21/2005&orggroup=27&effectivedate=4/22/2005&taskid=2
&comments=This is
to be
removed 
immediately&selected_taskid=2&orgfilterDISP=&taskname=Recall Meat
&equation=&tasktypeid=283&successRedirectAction=showTaskCompletionSummary
&estimatedduration=2:00&org=545";
</script>

Answer 1

使用encodeURIComponent：

var realQueryString = encodeURIComponent(queryString);

Answer 2

str = str.replace（/ \ s / g，“％20”）;

Answer 3

使用

Javascript url decode, encode

/**
*
*  URL encode / decode
*  http://www.webtoolkit.info/
*
**/

var Url = {

    // public method for url encoding
    encode : function (string) {
        return escape(this._utf8_encode(string));
    },

    // public method for url decoding
    decode : function (string) {
        return this._utf8_decode(unescape(string));
    },

    // private method for UTF-8 encoding
    _utf8_encode : function (string) {
        string = string.replace(/\r\n/g,"\n");
        var utftext = "";

        for (var n = 0; n < string.length; n++) {

            var c = string.charCodeAt(n);

            if (c < 128) {
                utftext += String.fromCharCode(c);
            }
            else if((c > 127) && (c < 2048)) {
                utftext += String.fromCharCode((c >> 6) | 192);
                utftext += String.fromCharCode((c & 63) | 128);
            }
            else {
                utftext += String.fromCharCode((c >> 12) | 224);
                utftext += String.fromCharCode(((c >> 6) & 63) | 128);
                utftext += String.fromCharCode((c & 63) | 128);
            }

        }

        return utftext;
    },

    // private method for UTF-8 decoding
    _utf8_decode : function (utftext) {
        var string = "";
        var i = 0;
        var c = c1 = c2 = 0;

        while ( i < utftext.length ) {

            c = utftext.charCodeAt(i);

            if (c < 128) {
                string += String.fromCharCode(c);
                i++;
            }
            else if((c > 191) && (c < 224)) {
                c2 = utftext.charCodeAt(i+1);
                string += String.fromCharCode(((c & 31) << 6) | (c2 & 63));
                i += 2;
            }
            else {
                c2 = utftext.charCodeAt(i+1);
                c3 = utftext.charCodeAt(i+2);
                string += String.fromCharCode(((c & 15) << 12) | ((c2 & 63) << 6) | (c3 & 63));
                i += 3;
            }

        }

        return string;
    }

}

格式化POST URL时出错，因为参数中有空格

3 个答案: