从URL创建一个JavaScript数组

时间:2010-11-28 16:30:16

标签: javascript url

我需要从URL创建一个Javascript数组,例如:

转过来:

http://maps.google.com/maps/api/staticmap?center=Baker Street 221b, London&size=450x450&markers=Baker Street 221b, London&sensor=false

类似于:

array['center'] = Baker Street 221b, London
array['size'] = 450x450
// and so on...

我需要使这个序列化/反序列化双向工作(url到数组和数组到url的部分)。是否有一些内置函数可以做到这一点?

提前致谢!

4 个答案:

答案 0 :(得分:46)

指向数组的网址:(改编自my answer here

function URLToArray(url) {
    var request = {};
    var pairs = url.substring(url.indexOf('?') + 1).split('&');
    for (var i = 0; i < pairs.length; i++) {
        if(!pairs[i])
            continue;
        var pair = pairs[i].split('=');
        request[decodeURIComponent(pair[0])] = decodeURIComponent(pair[1]);
     }
     return request;
}

数组到网址:

function ArrayToURL(array) {
  var pairs = [];
  for (var key in array)
    if (array.hasOwnProperty(key))

      pairs.push(encodeURIComponent(key) + '=' + encodeURIComponent(array[key]));
  return pairs.join('&');
}

答案 1 :(得分:2)

当url字符串有elem [] = 23&amp; elem [] = 56时,上面的函数URLToArray不起作用.. 请参阅下面的改编功能...希望它正常工作 - 不是100%经过测试

function URLToArray(url) {
        var request = {};
        var arr = [];
        var pairs = url.substring(url.indexOf('?') + 1).split('&');
        for (var i = 0; i < pairs.length; i++) {
          var pair = pairs[i].split('=');

          //check we have an array here - add array numeric indexes so the key elem[] is not identical.
          if(endsWith(decodeURIComponent(pair[0]), '[]') ) {
              var arrName = decodeURIComponent(pair[0]).substring(0, decodeURIComponent(pair[0]).length - 2);
              if(!(arrName in arr)) {
                  arr.push(arrName);
                  arr[arrName] = [];
              }

              arr[arrName].push(decodeURIComponent(pair[1]));
              request[arrName] = arr[arrName];
          } else {
            request[decodeURIComponent(pair[0])] = decodeURIComponent(pair[1]);
          }
        }
        return request;
    }

其中endWith取自here

function endsWith(str, suffix) {
    return str.indexOf(suffix, str.length - suffix.length) !== -1;
}

答案 2 :(得分:0)

 /**
 * (C)VIOLONIX inc.
 * Parser for make multidim array from
 * foo[]=any&foo[]=boy, or foo[0][kids]=any&foo[1][kids]=boy
 * result: foo=[[any],[boy]] or foo=[kids:[any],kids:[boy]]
 */    

 var URLToArray = function(url){
        function parse_mdim(name, val, data){
            let params = name.match(/(\[\])|(\[.+?\])/g);
            if(!params)params = new Array();
            let tg_id = name.split('[')[0];


            if(!(tg_id in data)) data[tg_id] = [];
            var prev_data = data[tg_id];

            for(var i=0;i<params.length;i++){
                if(params[i]!='[]'){
                    let tparam = params[i].match(/\[(.+)\]/i)[1];
                    if(!(tparam in prev_data)) prev_data[tparam] = [];
                    prev_data = prev_data[tparam];
                }else{
                    prev_data.push([]);
                    prev_data = prev_data[prev_data.length-1];
                }

            }
            prev_data.push(val);

        }


        var request = {};
        var arr = [];
        var pairs = url.substring(url.indexOf('?') + 1).split('&');

        for (var i = 0; i < pairs.length; i++) {
           var pair = pairs[i].split('=');
           if(decodeURIComponent(pair[0]).indexOf('[')!=-1)
               parse_mdim(decodeURIComponent(pair[0]), decodeURIComponent(pair[1]), request);
           else 
               request[decodeURIComponent(pair[0])] = decodeURIComponent(pair[1]);
         }

        //To-do here check array and simplifity it: if parameter end with one index in array replace it by value [0]    

         return request;

    }

答案 3 :(得分:-2)