据我所知，没有原生的解决方案。Dojo有一个内置的反序列化方法(如果您碰巧使用了该框架)。

否则，你可以自己简单地实现它:

function unserialize(str) {
  str = decodeURIComponent(str);
  var chunks = str.split('&'),
      obj = {};
  for(var c=0; c < chunks.length; c++) {
    var split = chunks[c].split('=', 2);
    obj[split[0]] = split[1];
  }
  return obj;
}

编辑:添加decodeURIComponent()

我也遇到了同样的问题，尝试了这里的解决方案，但没有一个真正有效，因为我在URL参数中有数组，像这样:

?param[]=5&param[]=8&othr_param=abc&param[]=string

所以我最终写了我自己的JS函数，它使一个数组的参数在URI:

/**
 * Creates an object from URL encoded data
 */
var createObjFromURI = function() {
    var uri = decodeURI(location.search.substr(1));
    var chunks = uri.split('&');
    var params = Object();

    for (var i=0; i < chunks.length ; i++) {
        var chunk = chunks[i].split('=');
        if(chunk[0].search("\\[\\]") !== -1) {
            if( typeof params[chunk[0]] === 'undefined' ) {
                params[chunk[0]] = [chunk[1]];

            } else {
                params[chunk[0]].push(chunk[1]);
            }


        } else {
            params[chunk[0]] = chunk[1];
        }
    }

    return params;
}

使用phpjs

function parse_str(str, array) {
  //       discuss at: http://phpjs.org/functions/parse_str/
  //      original by: Cagri Ekin
  //      improved by: Michael White (http://getsprink.com)
  //      improved by: Jack
  //      improved by: Brett Zamir (http://brett-zamir.me)
  //      bugfixed by: Onno Marsman
  //      bugfixed by: Brett Zamir (http://brett-zamir.me)
  //      bugfixed by: stag019
  //      bugfixed by: Brett Zamir (http://brett-zamir.me)
  //      bugfixed by: MIO_KODUKI (http://mio-koduki.blogspot.com/)
  // reimplemented by: stag019
  //         input by: Dreamer
  //         input by: Zaide (http://zaidesthings.com/)
  //         input by: David Pesta (http://davidpesta.com/)
  //         input by: jeicquest
  //             note: When no argument is specified, will put variables in global scope.
  //             note: When a particular argument has been passed, and the returned value is different parse_str of PHP. For example, a=b=c&d====c
  //             test: skip
  //        example 1: var arr = {};
  //        example 1: parse_str('first=foo&second=bar', arr);
  //        example 1: $result = arr
  //        returns 1: { first: 'foo', second: 'bar' }
  //        example 2: var arr = {};
  //        example 2: parse_str('str_a=Jack+and+Jill+didn%27t+see+the+well.', arr);
  //        example 2: $result = arr
  //        returns 2: { str_a: "Jack and Jill didn't see the well." }
  //        example 3: var abc = {3:'a'};
  //        example 3: parse_str('abc[a][b]["c"]=def&abc[q]=t+5');
  //        returns 3: {"3":"a","a":{"b":{"c":"def"}},"q":"t 5"}

  var strArr = String(str)
    .replace(/^&/, '')
    .replace(/&$/, '')
    .split('&'),
    sal = strArr.length,
    i, j, ct, p, lastObj, obj, lastIter, undef, chr, tmp, key, value,
    postLeftBracketPos, keys, keysLen,
    fixStr = function(str) {
      return decodeURIComponent(str.replace(/\+/g, '%20'));
    };

  if (!array) {
    array = this.window;
  }

  for (i = 0; i < sal; i++) {
    tmp = strArr[i].split('=');
    key = fixStr(tmp[0]);
    value = (tmp.length < 2) ? '' : fixStr(tmp[1]);

    while (key.charAt(0) === ' ') {
      key = key.slice(1);
    }
    if (key.indexOf('\x00') > -1) {
      key = key.slice(0, key.indexOf('\x00'));
    }
    if (key && key.charAt(0) !== '[') {
      keys = [];
      postLeftBracketPos = 0;
      for (j = 0; j < key.length; j++) {
        if (key.charAt(j) === '[' && !postLeftBracketPos) {
          postLeftBracketPos = j + 1;
        } else if (key.charAt(j) === ']') {
          if (postLeftBracketPos) {
            if (!keys.length) {
              keys.push(key.slice(0, postLeftBracketPos - 1));
            }
            keys.push(key.substr(postLeftBracketPos, j - postLeftBracketPos));
            postLeftBracketPos = 0;
            if (key.charAt(j + 1) !== '[') {
              break;
            }
          }
        }
      }
      if (!keys.length) {
        keys = [key];
      }
      for (j = 0; j < keys[0].length; j++) {
        chr = keys[0].charAt(j);
        if (chr === ' ' || chr === '.' || chr === '[') {
          keys[0] = keys[0].substr(0, j) + '_' + keys[0].substr(j + 1);
        }
        if (chr === '[') {
          break;
        }
      }

      obj = array;
      for (j = 0, keysLen = keys.length; j < keysLen; j++) {
        key = keys[j].replace(/^['"]/, '')
          .replace(/['"]$/, '');
        lastIter = j !== keys.length - 1;
        lastObj = obj;
        if ((key !== '' && key !== ' ') || j === 0) {
          if (obj[key] === undef) {
            obj[key] = {};
          }
          obj = obj[key];
        } else { // To insert new dimension
          ct = -1;
          for (p in obj) {
            if (obj.hasOwnProperty(p)) {
              if (+p > ct && p.match(/^\d+$/g)) {
                ct = +p;
              }
            }
          }
          key = ct + 1;
        }
      }
      lastObj[key] = value;
    }
  }
}

另一种基于URLSearchParams最新标准的解决方案(https://developer.mozilla.org/en-US/docs/Web/API/URLSearchParams)

function getQueryParamsObject() {
  const searchParams = new URLSearchParams(location.search.slice(1));
  return searchParams
    ? _.fromPairs(Array.from(searchParams.entries()))
    : {};
}

请注意，这个解决方案是利用

Array.from (https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/from)

和lodash的_. fropairs (https://lodash.com/docs#fromPairs)，以便简单。

因为您可以访问searchParams.entries()迭代器，所以创建一个更兼容的解决方案应该很容易。

许多其他的解决方案没有考虑到边界情况。

这个可以处理

空键a=1&b=2& 空值a=1&b 空值a=1&b= 未编码的等号a=1&b=2=3=4

  decodeQueryString: qs => {
    // expects qs to not have a ?
    // return if empty qs
    if (qs === '') return {};
    return qs.split('&').reduce((acc, pair) => {
      // skip no param at all a=1&b=2&
      if (pair.length === 0) return acc;
      const parts = pair.split('=');
      // fix params without value
      if (parts.length === 1) parts[1] = '';
      // for value handle multiple unencoded = signs
      const key = decodeURIComponent(parts[0]);
      const value = decodeURIComponent(parts.slice(1).join('='));
      acc[key] = value;
      return acc;
    }, {});
  },

一个简单的答案是内置本地Node模块。(没有第三方npm模块)

querystring模块提供了解析和格式化URL查询字符串的实用程序。可以使用以下方法访问:

const querystring = require('querystring');

const body = "abc=foo&def=%5Basf%5D&xyz=5"
const parseJSON = querystring.parse(body);
console.log(parseJSON);