398

在使用 AJAX 调用的 Web 应用程序中,我需要提交一个请求,但在 URL 的末尾添加一个参数,例如:

原网址:

http://server/myapp.php?id=10

结果网址:

http://server/myapp.php?id=10 &enabled=true

寻找一个 JavaScript 函数来解析 URL 并查看每个参数,然后添加新参数或更新值(如果已存在)。

4

35 回答 35

500

您可以使用以下之一:

例子:

var url = new URL("http://foo.bar/?x=1&y=2");

// If your expected result is "http://foo.bar/?x=1&y=2&x=42"
url.searchParams.append('x', 42);

// If your expected result is "http://foo.bar/?x=42&y=2"
url.searchParams.set('x', 42);
于 2017-05-24T14:12:24.287 回答
232

您需要调整的基本实现如下所示:

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;
}

这大约是基于正则表达式或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


我为完成而进行基准测试的慢正则表达式方法(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;
}
于 2009-01-28T09:50:48.897 回答
147
const urlParams = new URLSearchParams(window.location.search);

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

window.location.search = urlParams;

.set 第一个参数是键,第二个是值。

于 2019-07-08T10:34:10.457 回答
66

谢谢大家的贡献。我使用了 anakata代码并进行了修改,以包括 url 中根本没有查询字符串的情况。希望这会有所帮助。

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('&');
        }
    }
于 2009-12-16T21:36:29.880 回答
66

这是非常简单的解决方案。它不控制参数的存在,也不改变现有值。它将您的参数添加到 end,因此您可以在后端代码中获取最新值。

function addParameterToURL(param){
    _url = location.href;
    _url += (_url.split('?')[1] ? '&':'?') + param;
    return _url;
}
于 2011-01-26T23:05:45.047 回答
35

这是一个大大简化的版本,权衡易读性和更少的代码行而不是微优化的性能(实际上,我们谈论的是几毫秒的差异......由于它的性质(在当前文档的位置上运行) ),这很可能会在页面上运行一次)。

/**
* 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;
};

然后你会像这样使用它:

document.location.pathname + addUrlParam(document.location.search, 'foo', 'bar');
于 2010-04-13T13:45:50.550 回答
24

类中有一个内置函数URL,您可以使用它来轻松处理查询字符串键/值参数:

const url = new URL(window.location.href);
// url.searchParams has several function, we just use `set` function
// to set a value, if you just want to append without replacing value
// let use `append` function

url.searchParams.set('key', 'value');

console.log(url.search) // <== '?key=value'

// if window.location.href has already some qs params this `set` function
// modify or append key/value in it

有关更多信息searchParams functions

URLIE不支持,请检查兼容性

于 2021-05-23T21:32:47.510 回答
22

/**
* Add a URL parameter 
* @param {string} url 
* @param {string} param the key to set
* @param {string} value 
*/
var addParam = function(url, param, value) {
   param = encodeURIComponent(param);
   var a = document.createElement('a');
   param += (value ? "=" + encodeURIComponent(value) : ""); 
   a.href = url;
   a.search += (a.search ? "&" : "") + param;
   return a.href;
}

/**
* Add a URL parameter (or modify if already exists)
* @param {string} url 
* @param {string} param the key to set
* @param {string} value 
*/
var addOrReplaceParam = function(url, param, value) {
   param = encodeURIComponent(param);
   var r = "([&?]|&amp;)" + param + "\\b(?:=(?:[^&#]*))*";
   var a = document.createElement('a');
   var regex = new RegExp(r);
   var str = param + (value ? "=" + encodeURIComponent(value) : ""); 
   a.href = url;
   var q = a.search.replace(regex, "$1"+str);
   if (q === a.search) {
      a.search += (a.search ? "&" : "") + str;
   } else {
      a.search = q;
   }
   return a.href;
}

url = "http://www.example.com#hashme";
newurl = addParam(url, "ciao", "1");
alert(newurl);

请注意,参数应在附加到查询字符串之前进行编码。

http://jsfiddle.net/48z7z4kx/

于 2013-01-22T23:29:10.253 回答
20

我有一个这样做的“类”,它是:

function QS(){
    this.qs = {};
    var s = location.search.replace( /^\?|#.*$/g, '' );
    if( s ) {
        var qsParts = s.split('&');
        var i, nv;
        for (i = 0; i < qsParts.length; i++) {
            nv = qsParts[i].split('=');
            this.qs[nv[0]] = nv[1];
        }
    }
}

QS.prototype.add = function( name, value ) {
    if( arguments.length == 1 && arguments[0].constructor == Object ) {
        this.addMany( arguments[0] );
        return;
    }
    this.qs[name] = value;
}

QS.prototype.addMany = function( newValues ) {
    for( nv in newValues ) {
        this.qs[nv] = newValues[nv];
    }
}

QS.prototype.remove = function( name ) {
    if( arguments.length == 1 && arguments[0].constructor == Array ) {
        this.removeMany( arguments[0] );
        return;
    }
    delete this.qs[name];
}

QS.prototype.removeMany = function( deleteNames ) {
    var i;
    for( i = 0; i < deleteNames.length; i++ ) {
        delete this.qs[deleteNames[i]];
    }
}

QS.prototype.getQueryString = function() {
    var nv, q = [];
    for( nv in this.qs ) {
        q[q.length] = nv+'='+this.qs[nv];
    }
    return q.join( '&' );
}

QS.prototype.toString = QS.prototype.getQueryString;

//examples
//instantiation
var qs = new QS;
alert( qs );

//add a sinle name/value
qs.add( 'new', 'true' );
alert( qs );

//add multiple key/values
qs.add( { x: 'X', y: 'Y' } );
alert( qs );

//remove single key
qs.remove( 'new' )
alert( qs );

//remove multiple keys
qs.remove( ['x', 'bogus'] )
alert( qs );

我已经重写了 toString 方法,因此不需要调用 QS::getQueryString,您可以使用 QS::toString,或者,正如我在示例中所做的那样,只依赖被强制转换为字符串的对象。

于 2009-01-28T10:11:28.920 回答
12

如果你有一个带有 url 的字符串,你想用一个参数来装饰,你可以试试这个:

urlstring += ( urlstring.match( /[\?]/g ) ? '&' : '?' ) + 'param=value';

这意味着什么?将是参数的前缀,但如果你已经有了in urlstring, than &将是前缀。

encodeURI( paramvariable )如果您没有硬编码参数,我也建议您这样做,但它在paramvariable; 或者如果你有有趣的角色。

有关该函数的用法,请参阅javascript URL 编码。encodeURI

于 2016-03-17T17:54:32.673 回答
9

这是添加查询参数的简单方法:

const query = new URLSearchParams(window.location.search);
query.append("enabled", "true");

仅此而已。

请注意支持规格

于 2017-08-01T01:39:59.727 回答
8

有时我们?在最后的 URL 中看到,我发现了一些生成结果为file.php?&foo=bar. 我想出了我自己的解决方案,完全按照我的意愿工作!

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

注意:location.origin在 IE 中不起作用,这里是它的修复

于 2015-04-16T21:54:08.890 回答
7

以下功能将帮助您在 URL 中添加、更新和删除参数。

//例子1和

var myURL = '/search';

myURL = updateUrl(myURL,'location','california');
console.log('added location...' + myURL);
//added location.../search?location=california

myURL = updateUrl(myURL,'location','new york');
console.log('updated location...' + myURL);
//updated location.../search?location=new%20york

myURL = updateUrl(myURL,'location');
console.log('removed location...' + myURL);
//removed location.../search

//例子2

var myURL = '/search?category=mobile';

myURL = updateUrl(myURL,'location','california');
console.log('added location...' + myURL);
//added location.../search?category=mobile&location=california

myURL = updateUrl(myURL,'location','new york');
console.log('updated location...' + myURL);
//updated location.../search?category=mobile&location=new%20york

myURL = updateUrl(myURL,'location');
console.log('removed location...' + myURL);
//removed location.../search?category=mobile

//例子3

var myURL = '/search?location=texas';

myURL = updateUrl(myURL,'location','california');
console.log('added location...' + myURL);
//added location.../search?location=california

myURL = updateUrl(myURL,'location','new york');
console.log('updated location...' + myURL);
//updated location.../search?location=new%20york

myURL = updateUrl(myURL,'location');
console.log('removed location...' + myURL);
//removed location.../search

//例子4

var myURL = '/search?category=mobile&location=texas';

myURL = updateUrl(myURL,'location','california');
console.log('added location...' + myURL);
//added location.../search?category=mobile&location=california

myURL = updateUrl(myURL,'location','new york');
console.log('updated location...' + myURL);
//updated location.../search?category=mobile&location=new%20york

myURL = updateUrl(myURL,'location');
console.log('removed location...' + myURL);
//removed location.../search?category=mobile

//例子5

var myURL = 'https://example.com/search?location=texas#fragment';

myURL = updateUrl(myURL,'location','california');
console.log('added location...' + myURL);
//added location.../search?location=california#fragment

myURL = updateUrl(myURL,'location','new york');
console.log('updated location...' + myURL);
//updated location.../search?location=new%20york#fragment

myURL = updateUrl(myURL,'location');
console.log('removed location...' + myURL);
//removed location.../search#fragment

这是功能。

function updateUrl(url,key,value){
      if(value!==undefined){
        value = encodeURI(value);
      }
      var hashIndex = url.indexOf("#")|0;
      if (hashIndex === -1) hashIndex = url.length|0;
      var urls = url.substring(0, hashIndex).split('?');
      var baseUrl = urls[0];
      var parameters = '';
      var outPara = {};
      if(urls.length>1){
          parameters = urls[1];
      }
      if(parameters!==''){
        parameters = parameters.split('&');
        for(k in parameters){
          var keyVal = parameters[k];
          keyVal = keyVal.split('=');
          var ekey = keyVal[0];
          var evalue = '';
          if(keyVal.length>1){
              evalue = keyVal[1];
          }
          outPara[ekey] = evalue;
        }
      }

      if(value!==undefined){
        outPara[key] = value;
      }else{
        delete outPara[key];
      }
      parameters = [];
      for(var k in outPara){
        parameters.push(k + '=' + outPara[k]);
      }

      var finalUrl = baseUrl;

      if(parameters.length>0){
        finalUrl += '?' + parameters.join('&'); 
      }

      return finalUrl + url.substring(hashIndex); 
  }
于 2016-05-26T11:28:40.237 回答
6

这是我自己的尝试,但我会使用 annakata 的答案,因为它看起来更干净:

function AddUrlParameter(sourceUrl, parameterName, parameterValue, replaceDuplicates)
{
    if ((sourceUrl == null) || (sourceUrl.length == 0)) sourceUrl = document.location.href;
    var urlParts = sourceUrl.split("?");
    var newQueryString = "";
    if (urlParts.length > 1)
    {
        var parameters = urlParts[1].split("&");
        for (var i=0; (i < parameters.length); i++)
        {
            var parameterParts = parameters[i].split("=");
            if (!(replaceDuplicates && parameterParts[0] == parameterName))
            {
                if (newQueryString == "")
                    newQueryString = "?";
                else
                    newQueryString += "&";
                newQueryString += parameterParts[0] + "=" + parameterParts[1];
            }
        }
    }
    if (newQueryString == "")
        newQueryString = "?";
    else
        newQueryString += "&";
    newQueryString += parameterName + "=" + parameterValue;

    return urlParts[0] + newQueryString;
}

另外,我在 stackoverflow 上的另一篇文章中找到了这个 jQuery 插件,如果你需要更多的灵活性,你可以使用它: http: //plugins.jquery.com/project/query-object

我认为代码将是(尚未测试):

return $.query.parse(sourceUrl).set(parameterName, parameterValue).toString();
于 2009-01-28T10:04:09.027 回答
6

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

javascript 中的 Uri 和查询字符串操作。

这个项目包含了 Steven Levithan 的优秀 parseUri 正则表达式库。您可以安全地解析各种形状和大小的 URL,无论它们是无效的还是可怕的。

于 2013-05-23T19:23:06.703 回答
6

试试这个。

// uses the URL class
function setParam(key, value) {
            let url = new URL(window.document.location);
            let params = new URLSearchParams(url.search.slice(1));

            if (params.has(key)) {
                params.set(key, value);
            }else {
                params.append(key, value);
            }
        }
于 2019-10-20T08:52:06.340 回答
5

添加到@Vianney 的答案https://stackoverflow.com/a/44160941/6609678

我们可以在node中导入Built-in URL模块如下

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

于 2019-03-11T07:46:54.880 回答
5

它处理这样的 URL:

  • 空的
  • 没有任何参数
  • 已经有一些参数
  • 末尾有?,但同时没有任何参数

它不处理这样的 URL:

  • 带有片段标识符(即哈希,#)
  • 如果 URL 已经有必需的查询参数(那么会有重复)

工作于:

  • 铬 32+
  • 火狐 26+
  • Safari 7.1+
function appendQueryParameter(url, name, value) {
    if (url.length === 0) {
        return;
    }

    let rawURL = url;

    // URL with `?` at the end and without query parameters
    // leads to incorrect result.
    if (rawURL.charAt(rawURL.length - 1) === "?") {
        rawURL = rawURL.slice(0, rawURL.length - 1);
    }

    const parsedURL = new URL(rawURL);
    let parameters = parsedURL.search;

    parameters += (parameters.length === 0) ? "?" : "&";
    parameters = (parameters + name + "=" + value);

    return (parsedURL.origin + parsedURL.pathname + parameters);
}

带有 ES6 模板字符串的版本。

工作于:

  • 铬 41+
  • 火狐 32+
  • Safari 9.1+
function appendQueryParameter(url, name, value) {
    if (url.length === 0) {
        return;
    }

    let rawURL = url;

    // URL with `?` at the end and without query parameters
    // leads to incorrect result.
    if (rawURL.charAt(rawURL.length - 1) === "?") {
        rawURL = rawURL.slice(0, rawURL.length - 1);
    }

    const parsedURL = new URL(rawURL);
    let parameters = parsedURL.search;

    parameters += (parameters.length === 0) ? "?" : "&";
    parameters = `${parameters}${name}=${value}`;

    return `${parsedURL.origin}${parsedURL.pathname}${parameters}`;
}
于 2020-07-15T13:55:31.580 回答
4

此解决方案使用更新的搜索参数更新窗口的当前 URL,而无需实际重新加载页面。这种方法在 PWA/SPA 上下文中很有用。

function setURLSearchParam(key, value) {
  const url = new URL(window.location.href);
  url.searchParams.set(key, value);
  window.history.pushState({ path: url.href }, '', url.href);
}
于 2021-11-20T20:35:42.330 回答
3

Vianney Bajart 的回答是正确的;但是,只有当您拥有包含端口、主机、路径和查询的完整 URL 时, URL才会起作用:

new URL('http://server/myapp.php?id=10&enabled=true')

并且URLSearchParams仅在您仅传递查询字符串时才有效:

new URLSearchParams('?id=10&enabled=true')

如果您有一个不完整或相对的 URL 并且不关心基本 URL,您可以只拆分?以获取查询字符串并稍后加入,如下所示:

function setUrlParams(url, key, value) {
  url = url.split('?');
  usp = new URLSearchParams(url[1]);
  usp.set(key, value);
  url[1] = usp.toString();
  return url.join('?');
}

let url = 'myapp.php?id=10';
url = setUrlParams(url, 'enabled', true);  // url = 'myapp.php?id=10&enabled=true'
url = setUrlParams(url, 'id', 11);         // url = 'myapp.php?id=11&enabled=true'

与 Internet Explorer 不兼容。

于 2018-05-28T19:42:28.097 回答
2

我喜欢 Mehmet Fatih Yıldız 的回答,即使他没有回答整个问题。

在与他的回答相同的行中,我使用以下代码:

“它不控制参数的存在,也不会改变现有值。它将您的参数添加到末尾”

  /** 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;
  }

和测试仪:

  /*
  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');

  }//*/
于 2011-08-31T10:33:25.330 回答
2

我会使用这个小而完整的库来处理 js 中的 url:

https://github.com/Mikhus/jsurl

于 2013-07-02T20:33:08.247 回答
2

当涉及到像 Node.js 这样的服务器端的一些基本 url 参数添加或更新时,这就是我使用的。

咖啡脚本:

###
    @method addUrlParam Adds parameter to a given url. If the parameter already exists in the url is being replaced.
    @param {string} url
    @param {string} key Parameter's key
    @param {string} value Parameter's value
    @returns {string} new url containing the parameter
###
addUrlParam = (url, key, value) ->
    newParam = key+"="+value
    result = url.replace(new RegExp('(&|\\?)' + key + '=[^\&|#]*'), '$1' + newParam)
    if result is url
        result = if url.indexOf('?') != -1 then url.split('?')[0] + '?' + newParam + '&' + url.split('?')[1]
    else if url.indexOf('#') != -1 then url.split('#')[0] + '?' + newParam + '#' + url.split('#')[1]
    else url + '?' + newParam
    return result

JavaScript:

function addUrlParam(url, key, value) {
    var newParam = key+"="+value;
    var result = url.replace(new RegExp("(&|\\?)"+key+"=[^\&|#]*"), '$1' + newParam);
    if (result === url) { 
        result = (url.indexOf("?") != -1 ? url.split("?")[0]+"?"+newParam+"&"+url.split("?")[1] 
           : (url.indexOf("#") != -1 ? url.split("#")[0]+"?"+newParam+"#"+ url.split("#")[1] 
              : url+'?'+newParam));
    }
    return result;
}

var url = "http://www.example.com?foo=bar&ciao=3&doom=5#hashme";
result1.innerHTML = addUrlParam(url, "ciao", "1");
<p id="result1"></p>

于 2015-07-06T21:57:14.263 回答
2

最简单的解决方案,无论您是否已经有标签都可以使用,并自动将其删除,因此它不会继续添加相同的标签,玩得开心

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;
}

然后

changeURL("i=updated");
于 2017-07-13T10:17:53.823 回答
1

如果您在链接或其他地方弄乱了网址,您可能还必须考虑哈希。这是一个相当简单易懂的解决方案。可能不是最快的 ,因为它使用了正则表达式……但在 99.999% 的情况下,差异真的无关紧要!

function addQueryParam( url, key, val ){
    var parts = url.match(/([^?#]+)(\?[^#]*)?(\#.*)?/);
    var url = parts[1];
    var qs = parts[2] || '';
    var hash = parts[3] || '';

    if ( !qs ) {
        return url + '?' + key + '=' + encodeURIComponent( val ) + hash;
    } else {
        var qs_parts = qs.substr(1).split("&");
        var i;
        for (i=0;i<qs_parts.length;i++) {
            var qs_pair = qs_parts[i].split("=");
            if ( qs_pair[0] == key ){
                qs_parts[ i ] = key + '=' + encodeURIComponent( val );
                break;
            }
        }
        if ( i == qs_parts.length ){
            qs_parts.push( key + '=' + encodeURIComponent( val ) );
        }
        return url + '?' + qs_parts.join('&') + hash;
    }
}
于 2013-02-21T19:42:59.050 回答
0

好的,这里我比较两个函数,一个是我自己做的(regExp),另一个是(an​​nakata)做的。

拆分数组:

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;
}

测试案例:

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(当正则表达式仅使用测试并且不输入 .replace 函数时)它仍然比拆分慢......嗯。正则表达式有点慢但是......呃......

于 2009-01-28T12:07:52.583 回答
0

这就是我所做的。使用我的 editParams() 函数,您可以添加、删除或更改任何参数,然后使用内置的 replaceState() 函数更新 URL:

window.history.replaceState('object or string', 'Title', 'page.html' + editParams('enable', 'true'));


// background functions below:

// add/change/remove URL parameter
// use a value of false to remove parameter
// returns a url-style string
function editParams (key, value) {
  key = encodeURI(key);

  var params = getSearchParameters();

  if (Object.keys(params).length === 0) {
    if (value !== false)
      return '?' + key + '=' + encodeURI(value);
    else
      return '';
  }

  if (value !== false)
    params[key] = encodeURI(value);
  else
    delete params[key];

  if (Object.keys(params).length === 0)
    return '';

  return '?' + $.map(params, function (value, key) {
    return key + '=' + value;
  }).join('&');
}

// Get object/associative array of URL parameters
function getSearchParameters () {
  var prmstr = window.location.search.substr(1);
  return prmstr !== null && prmstr !== "" ? transformToAssocArray(prmstr) : {};
}

// convert parameters from url-style string to associative array
function transformToAssocArray (prmstr) {
  var params = {},
      prmarr = prmstr.split("&");

  for (var i = 0; i < prmarr.length; i++) {
    var tmparr = prmarr[i].split("=");
    params[tmparr[0]] = tmparr[1];
  }
  return params;
}
于 2014-06-18T18:13:34.863 回答
0

尽我所能告诉以上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

以下函数是我写的 update document.location.search。它将一个键/值对数组的数组作为参数,它将返回后者的修订版本,您可以随心所欲地做任何事情。我这样使用它:

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

如果当前网址是:

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

这会让你

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

功能

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;
}
于 2017-01-20T16:47:40.980 回答
0

试试
正则表达式,这么慢,于是:

var SetParamUrl = function(_k, _v) {// replace and add new parameters

    let arrParams = window.location.search !== '' ? decodeURIComponent(window.location.search.substr(1)).split('&').map(_v => _v.split('=')) : Array();
    let index = arrParams.findIndex((_v) => _v[0] === _k); 
    index = index !== -1 ? index : arrParams.length;
    _v === null ? arrParams = arrParams.filter((_v, _i) => _i != index) : arrParams[index] = [_k, _v];
    let _search = encodeURIComponent(arrParams.map(_v => _v.join('=')).join('&'));

    let newurl = window.location.protocol + "//" + window.location.host + window.location.pathname + (arrParams.length > 0 ? '?' +  _search : ''); 

    // window.location = newurl; //reload 

    if (history.pushState) { // without reload  
        window.history.pushState({path:newurl}, null, newurl);
    }

};

var GetParamUrl = function(_k) {// get parameter by key

    let sPageURL = decodeURIComponent(window.location.search.substr(1)),
        sURLVariables = sPageURL.split('&').map(_v => _v.split('='));
    let _result = sURLVariables.find(_v => _v[0] === _k);
    return _result[1];

};

例子:

        // https://some.com/some_path
        GetParamUrl('cat');//undefined
        SetParamUrl('cat', "strData");// https://some.com/some_path?cat=strData
        GetParamUrl('cat');//strData
        SetParamUrl('sotr', "strDataSort");// https://some.com/some_path?cat=strData&sotr=strDataSort
        GetParamUrl('sotr');//strDataSort
        SetParamUrl('cat', "strDataTwo");// https://some.com/some_path?cat=strDataTwo&sotr=strDataSort
        GetParamUrl('cat');//strDataTwo
        //remove param
        SetParamUrl('cat', null);// https://some.com/some_path?sotr=strDataSort
于 2017-08-27T17:34:39.727 回答
0

借助 JS 中的新成就,可以将查询参数添加到 URL:

var protocol = window.location.protocol,
    host = '//' + window.location.host,
    path = window.location.pathname,
    query = window.location.search;

var newUrl = protocol + host + path + query + (query ? '&' : '?') + 'param=1';

window.history.pushState({path:newUrl}, '' , newUrl);

也看到这种可能性Moziila URLSearchParams.append()

于 2018-02-02T11:14:10.423 回答
0

这适用于所有现代浏览器。

function insertParam(key,value) {
      if (history.pushState) {
          var newurl = window.location.protocol + "//" + window.location.host + window.location.pathname + '?' +key+'='+value;
          window.history.pushState({path:newurl},'',newurl);
      }
    }
于 2018-03-24T12:48:22.443 回答
0

重置所有查询字符串

var params = { params1:"val1", params2:"val2" };
let str = jQuery.param(params);

let uri = window.location.href.toString();
if (uri.indexOf("?") > 0)
   uri = uri.substring(0, uri.indexOf("?"));

console.log(uri+"?"+str);
//window.location.href = uri+"?"+str;
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

于 2019-07-10T07:05:58.307 回答
0

以下:

  • 合并重复的查询字符串参数
  • 适用于绝对和相对 URL
  • 在浏览器和节点中工作
/**
 * Adds query params to existing URLs (inc merging duplicates)
 * @param {string} url - src URL to modify
 * @param {object} params - key/value object of params to add
 * @returns {string} modified URL
 */
function addQueryParamsToUrl(url, params) {

    // if URL is relative, we'll need to add a fake base
    var fakeBase = !url.startsWith('http') ? 'http://fake-base.com' : undefined;
    var modifiedUrl = new URL(url || '', fakeBase);

    // add/update params
    Object.keys(params).forEach(function(key) {
        if (modifiedUrl.searchParams.has(key)) {
            modifiedUrl.searchParams.set(key, params[key]);
        }
        else {
            modifiedUrl.searchParams.append(key, params[key]);
        }
    });

    // return as string (remove fake base if present)
    return modifiedUrl.toString().replace(fakeBase, '');
}

例子:

// returns /guides?tag=api
addQueryParamsToUrl('/guides?tag=hardware', { tag:'api' })

// returns https://orcascan.com/guides?tag=api
addQueryParamsToUrl('https://orcascan.com/guides?tag=hardware', { tag: 'api' })
于 2022-01-05T19:57:34.530 回答
0
const params = new URLSearchParams(window.location.search);

params.delete(key)
window.history.replaceState({}, "", decodeURIComponent(`${window.location.pathname}?${params}`));
于 2022-03-06T03:44:42.260 回答
-2
var MyApp = new Class();

MyApp.extend({
    utility: {
        queryStringHelper: function (url) {
            var originalUrl = url;
            var newUrl = url;
            var finalUrl;
            var insertParam = function (key, value) {
                key = escape(key);
                value = escape(value);

                //The previous post had the substr strat from 1 in stead of 0!!!
                var kvp = newUrl.substr(0).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('=');
                }

                finalUrl = kvp.join('&');

                return finalUrl;
            };

            this.insertParameterToQueryString = insertParam;

            this.insertParams = function (keyValues) {
                for (var keyValue in keyValues[0]) {
                    var key = keyValue;
                    var value = keyValues[0][keyValue];
                    newUrl = insertParam(key, value);
                }
                return newUrl;
            };

            return this;
        }
    }
});
于 2012-09-26T20:10:22.140 回答