我添加我的解决方案，因为它支持相对url除了绝对url。在其他方面，它与顶部的答案相同，后者也使用Web API。

/**
 * updates a relative or absolute
 * by setting the search query with
 * the passed key and value.
 */
export const setQueryParam = (url, key, value) => {
  const dummyBaseUrl = 'https://dummy-base-url.com';
  const result = new URL(url, dummyBaseUrl);
  result.searchParams.set(key, value);
  return result.toString().replace(dummyBaseUrl, '');
};

还有人开玩笑说:

// some jest tests
describe('setQueryParams', () => {
  it('sets param on relative url with base path', () => {
    // act
    const actual = setQueryParam(
      '/', 'ref', 'some-value',
    );
    // assert
    expect(actual).toEqual('/?ref=some-value');
  });
  it('sets param on relative url with no path', () => {
    // act
    const actual = setQueryParam(
      '', 'ref', 'some-value',
    );
    // assert
    expect(actual).toEqual('/?ref=some-value');
  });
  it('sets param on relative url with some path', () => {
    // act
    const actual = setQueryParam(
      '/some-path', 'ref', 'some-value',
    );
    // assert
    expect(actual).toEqual('/some-path?ref=some-value');
  });
  it('overwrites existing param', () => {
    // act
    const actual = setQueryParam(
      '/?ref=prev-value', 'ref', 'some-value',
    );
    // assert
    expect(actual).toEqual('/?ref=some-value');
  });
  it('sets param while another param exists', () => {
    // act
    const actual = setQueryParam(
      '/?other-param=other-value', 'ref', 'some-value',
    );
    // assert
    expect(actual).toEqual('/?other-param=other-value&ref=some-value');
  });
  it('honors existing base url', () => {
    // act
    const actual = setQueryParam(
      'https://base.com', 'ref', 'some-value',
    );
    // assert
    expect(actual).toEqual('https://base.com/?ref=some-value');
  });
  it('honors existing base url with some path', () => {
    // act
    const actual = setQueryParam(
      'https://base.com/some-path', 'ref', 'some-value',
    );
    // assert
    expect(actual).toEqual('https://base.com/some-path?ref=some-value');
  });
});

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');

  }//*/

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…

最简单的解决方案，工作，如果你已经有一个标签或没有，并自动删除它，所以它不会一直添加相等的标签，有乐趣

function changeURL(tag)
{
if(window.location.href.indexOf("?") > -1) {
    if(window.location.href.indexOf("&"+tag) > -1){

        var url = window.location.href.replace("&"+tag,"")+"&"+tag;
    }
    else
    {
        var url = window.location.href+"&"+tag;
    }
}else{
    if(window.location.href.indexOf("?"+tag) > -1){

        var url = window.location.href.replace("?"+tag,"")+"?"+tag;
    }
    else
    {
        var url = window.location.href+"?"+tag;
    }
}
  window.location = url;
}

THEN

changeURL("i=updated");

就我所知，上面的答案都没有解决查询字符串包含参数的情况，这些参数本身就是一个数组，因此会出现不止一次，例如:

http://example.com?sizes[]=a&sizes[]=b

下面的函数是我用来更新document.location.search的。它接受一个键/值对数组作为参数，它将返回一个修改后的版本，你可以做任何你想做的事情。我是这样使用的:

var newParams = [
    ['test','123'],
    ['best','456'],
    ['sizes[]','XXL']
];
var newUrl = document.location.pathname + insertParams(newParams);
history.replaceState('', '', newUrl);

如果当前url为:

http://example.com/index.php?test=replaceme&sizes [] = XL

这会让你

http://example.com/index.php?test=123&sizes[]=XL&尺寸[]=XXL&best=456

函数

function insertParams(params) {
    var result;
    var ii = params.length;
    var queryString = document.location.search.substr(1);
    var kvps = queryString ? queryString.split('&') : [];
    var kvp;
    var skipParams = [];
    var i = kvps.length;
    while (i--) {
        kvp = kvps[i].split('=');
        if (kvp[0].slice(-2) != '[]') {
            var ii = params.length;
            while (ii--) {
                if (params[ii][0] == kvp[0]) {
                    kvp[1] = params[ii][1];
                    kvps[i] = kvp.join('=');
                    skipParams.push(ii);
                }
            }
        }
    }
    var ii = params.length;
    while (ii--) {
        if (skipParams.indexOf(ii) === -1) {
            kvps.push(params[ii].join('='));
        }
    }
    result = kvps.length ? '?' + kvps.join('&') : '';
    return result;
}

我会用这个小而完整的库来处理js中的url:

https://github.com/Mikhus/jsurl