我发现$. string .deparam是最完整的预构建解决方案(可以做嵌套对象等)。查看文档。

首先你需要定义什么是get变量:

function getVar()
{
    this.length = 0;
    this.keys = [];
    this.push = function(key, value)
    {
        if(key=="") key = this.length++;
        this[key] = value;
        this.keys.push(key);
        return this[key];
    }
}

而不是直接读:

function urlElement()
{
    var thisPrototype = window.location;
    for(var prototypeI in thisPrototype) this[prototypeI] = thisPrototype[prototypeI];
    this.Variables = new getVar();
    if(!this.search) return this;
    var variables = this.search.replace(/\?/g,'').split('&');
    for(var varI=0; varI<variables.length; varI++)
    {
        var nameval = variables[varI].split('=');
        var name = nameval[0].replace(/\]/g,'').split('[');
        var pVariable = this.Variables;
        for(var nameI=0;nameI<name.length;nameI++)
        {
            if(name.length-1==nameI) pVariable.push(name[nameI],nameval[1]);
            else var pVariable = (typeof pVariable[name[nameI]] != 'object')? pVariable.push(name[nameI],new getVar()) : pVariable[name[nameI]];
        }
    }
}

并使用like:

var mlocation = new urlElement();
mlocation = mlocation.Variables;
for(var key=0;key<mlocation.keys.length;key++)
{
    console.log(key);
    console.log(mlocation[mlocation.keys[key]];
}

下面是我用的一个例子:

var params = {};
window.location.search.substring(1).split('&').forEach(function(pair) {
  pair = pair.split('=');
  if (pair[1] !== undefined) {
    var key = decodeURIComponent(pair[0]),
        val = decodeURIComponent(pair[1]),
        val = val ? val.replace(/\++/g,' ').trim() : '';

    if (key.length === 0) {
      return;
    }
    if (params[key] === undefined) {
      params[key] = val;
    }
    else {
      if ("function" !== typeof params[key].push) {
        params[key] = [params[key]];
      }
      params[key].push(val);
    }
  }
});
console.log(params);

基本用法。 ? = aa&b = bb 对象{a: "aa"， b: "bb"}

重复参数，例如。 ? = aa&b = bb&c = cc&c =土豆 对象{a: "aa"， b: "bb"， c: ["cc"，"potato"]}

钥匙不见了。 ? = aa&b = bb = cc 对象{a: "aa"， b: "bb"}

缺少值，例如。 = aa&b = bb&c ? 对象{a: "aa"， b: "bb"}

上述JSON/regex解决方案在这个古怪的url上抛出了一个语法错误: ? = aa&b = bb&c = & = dd&e 对象{a: "aa"， b: "bb"， c: ""}

/** * Parses and builds Object of URL query string. * @param {string} query The URL query string. * @return {!Object<string, string>} */ function parseQueryString(query) { if (!query) { return {}; } return (/^[?#]/.test(query) ? query.slice(1) : query) .split('&') .reduce((params, param) => { const item = param.split('='); const key = decodeURIComponent(item[0] || ''); const value = decodeURIComponent(item[1] || ''); if (key) { params[key] = value; } return params; }, {}); } console.log(parseQueryString('?v=MFa9pvnVe0w&ku=user&from=89&aw=1')) see log

一个班轮。干净简单。

const params = Object.fromEntries(new URLSearchParams(location.search));

对于您的具体情况，它将是:

const str = 'abc=foo&def=%5Basf%5D&xyz=5'; const params = Object.fromEntries(new URLSearchParams(str)); console.log (params);

ES6有一个非常简单而不正确的答案:

console.log(
  Object.fromEntries(new URLSearchParams(`abc=foo&def=%5Basf%5D&xyz=5`))
);

但是这一行代码不包括多个相同的键，你必须使用更复杂的东西:

function parseParams(params) {
  const output = [];
  const searchParams = new URLSearchParams(params);

  // Set will return only unique keys()
  new Set([...searchParams.keys()])
    .forEach(key => {
      output[key] = searchParams.getAll(key).length > 1 ?  
        searchParams.getAll(key) : // get multiple values
        searchParams.get(key); // get single value
    });

  return output;
}

console.log(
   parseParams('abc=foo&cars=Ford&cars=BMW&cars=Skoda&cars=Mercedes')
)

代码将生成如下结构:

[
  abc: "foo"
  cars: ["Ford", "BMW", "Skoda", "Mercedes"]
]