在一个使用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:

原始URL:

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

导致的网址:

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

寻找一个JavaScript函数,该函数解析URL并查看每个参数,然后添加新参数或更新已经存在的值。


当前回答

这是我自己的尝试,但我将使用annakata的答案,因为它看起来更清晰:

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,我从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

我认为代码应该是(还没有测试):

return $.query.parse(sourceUrl).set(parameterName, parameterValue).toString();

其他回答

以下功能将帮助您添加,更新和删除参数或从URL。

/ / example1and

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

/ / example2

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

/ /青年们

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

/ / example4

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

/ / example5

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

这是函数。

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); 
  }

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');

  }//*/

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

这将在所有现代浏览器中工作。

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);
      }
    }

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;
}