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;
}

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

感谢大家的贡献。我使用annakata代码并修改为也包括url中根本没有查询字符串的情况。 希望这能有所帮助。

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

        var kvp = document.location.search.substr(1).split('&');
        if (kvp == '') {
            document.location.search = '?' + key + '=' + value;
        }
        else {

            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
            document.location.search = kvp.join('&');
        }
    }

这是一个非常简化的版本，为了可读性和更少的代码行而不是微观优化的性能(我们说的是几毫秒的差异，实际上……由于它的性质(操作当前文档的位置)，这将很可能在一个页面上运行一次)。

/**
* Add a URL parameter (or changing it if it already exists)
* @param {search} string  this is typically document.location.search
* @param {key}    string  the key to set
* @param {val}    string  value 
*/
var addUrlParam = function(search, key, val){
  var newParam = key + '=' + val,
      params = '?' + newParam;

  // If the "search" string exists, then build params from it
  if (search) {
    // Try to replace an existance instance
    params = search.replace(new RegExp('([?&])' + key + '[^&]*'), '$1' + newParam);

    // If nothing was replaced, then add the new param to the end
    if (params === search) {
      params += '&' + newParam;
    }
  }

  return params;
};

然后你可以这样使用:

document.location.pathname + addUrlParam(document.location.search, 'foo', 'bar');

查看https://github.com/derek-watson/jsUri

Uri和javascript查询字符串操作。

这个项目结合了Steven Levithan的优秀parseUri正则表达式库。您可以安全地解析所有形状和大小的url，无论它们是多么无效或丑陋。

有时我们看到?在URL结尾，我找到了一些解决方案，生成的结果为file.php?&foo=bar。我想出了我自己的解决方案，以完美地工作!

location.origin + location.pathname + location.search + (location.search=='' ? '?' : '&') + 'lang=ar'

注意:位置。origin不能在IE中工作，这里是它的修复。