我会用这个小而完整的库来处理js中的url:

https://github.com/Mikhus/jsurl

你可以使用URLSearchParams

const urlParams = new URLSearchParams(window.location.search);

urlParams.set('order', 'date');

window.location.search = urlParams;

.第一个农业是关键，第二个是价值。

注意:这在任何版本的ie浏览器中都不支持(但在Edge中支持)

加入@Vianney的回答https://stackoverflow.com/a/44160941/6609678

我们可以导入node中的内置URL模块，如下所示

const { URL } = require('url');

例子:

Terminal $ node
> const { URL } = require('url');
undefined
> let url = new URL('', 'http://localhost:1989/v3/orders');
undefined
> url.href
'http://localhost:1989/v3/orders'
> let fetchAll=true, timePeriod = 30, b2b=false;
undefined
> url.href
'http://localhost:1989/v3/orders'
>  url.searchParams.append('fetchAll', fetchAll);
undefined
>  url.searchParams.append('timePeriod', timePeriod);
undefined
>  url.searchParams.append('b2b', b2b);
undefined
> url.href
'http://localhost:1989/v3/orders?fetchAll=true&timePeriod=30&b2b=false'
> url.toString()
'http://localhost:1989/v3/orders?fetchAll=true&timePeriod=30&b2b=false'

有用的链接:

https://developer.mozilla.org/en-US/docs/Web/API/URL https://developer.mozilla.org/en/docs/Web/API/URLSearchParams

好的，在这里我比较两个函数，一个由我自己(regExp)和另一个由(annakata)。

将数组:

function insertParam(key, value)
{
    key = escape(key); value = escape(value);

    var kvp = document.location.search.substr(1).split('&');

    var i=kvp.length; var x; while(i--) 
    {
        x = kvp[i].split('=');

        if (x[0]==key)
        {
                x[1] = value;
                kvp[i] = x.join('=');
                break;
        }
    }

    if(i<0) {kvp[kvp.length] = [key,value].join('=');}

    //this will reload the page, it's likely better to store this until finished
    return "&"+kvp.join('&'); 
}

正则表达式的方法:

function addParameter(param, value)
{
    var regexp = new RegExp("(\\?|\\&)" + param + "\\=([^\\&]*)(\\&|$)");
    if (regexp.test(document.location.search)) 
        return (document.location.search.toString().replace(regexp, function(a, b, c, d)
        {
                return (b + param + "=" + value + d);
        }));
    else 
        return document.location.search+ param + "=" + value;
}

测试用例:

time1=(new Date).getTime();
for (var i=0;i<10000;i++)
{
addParameter("test","test");
}
time2=(new Date).getTime();
for (var i=0;i<10000;i++)
{
insertParam("test","test");
}

time3=(new Date).getTime();

console.log((time2-time1)+" "+(time3-time2));

似乎即使使用最简单的解决方案(当regexp只使用test而不输入.replace函数时)，它仍然比分裂要慢…好。Regexp有点慢，但是…喔…

我喜欢穆罕穆德·法提赫·耶尔达兹的回答，即使他没有回答整个问题。

在他回答的同一行中，我使用了这样的代码:

“它不控制参数的存在，也不改变现有的值。它把你的参数加到最后"

  /** add a parameter at the end of the URL. Manage '?'/'&', but not the existing parameters.
   *  does escape the value (but not the key)
   */
  function addParameterToURL(_url,_key,_value){
      var param = _key+'='+escape(_value);

      var sep = '&';
      if (_url.indexOf('?') < 0) {
        sep = '?';
      } else {
        var lastChar=_url.slice(-1);
        if (lastChar == '&') sep='';
        if (lastChar == '?') sep='';
      }
      _url += sep + param;

      return _url;
  }

测试者:

  /*
  function addParameterToURL_TESTER_sub(_url,key,value){
    //log(_url);
    log(addParameterToURL(_url,key,value));
  }

  function addParameterToURL_TESTER(){
    log('-------------------');
    var _url ='www.google.com';
    addParameterToURL_TESTER_sub(_url,'key','value');
    addParameterToURL_TESTER_sub(_url,'key','Text Value');
    _url ='www.google.com?';
    addParameterToURL_TESTER_sub(_url,'key','value');
    _url ='www.google.com?A=B';
    addParameterToURL_TESTER_sub(_url,'key','value');
    _url ='www.google.com?A=B&';
    addParameterToURL_TESTER_sub(_url,'key','value');
    _url ='www.google.com?A=1&B=2';
    addParameterToURL_TESTER_sub(_url,'key','value');

  }//*/

你需要适应的基本实现是这样的:

function insertParam(key, value) {
    key = encodeURIComponent(key);
    value = encodeURIComponent(value);

    // kvp looks like ['key1=value1', 'key2=value2', ...]
    var kvp = document.location.search.substr(1).split('&');
    let i=0;

    for(; i<kvp.length; i++){
        if (kvp[i].startsWith(key + '=')) {
            let pair = kvp[i].split('=');
            pair[1] = value;
            kvp[i] = pair.join('=');
            break;
        }
    }

    if(i >= kvp.length){
        kvp[kvp.length] = [key,value].join('=');
    }

    // can return this or...
    let params = kvp.join('&');

    // reload page with new params
    document.location.search = params;
}

这大约是正则表达式或基于搜索的解决方案的两倍，但这完全取决于查询字符串的长度和任何匹配的索引

为了完成起见，我以慢速regex方法为基准(大约慢了150%)

function insertParam2(key,value)
{
    key = encodeURIComponent(key); value = encodeURIComponent(value);

    var s = document.location.search;
    var kvp = key+"="+value;

    var r = new RegExp("(&|\\?)"+key+"=[^\&]*");

    s = s.replace(r,"$1"+kvp);

    if(!RegExp.$1) {s += (s.length>0 ? '&' : '?') + kvp;};

    //again, do what you will here
    document.location.search = s;
}