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

试一试 正则表达式,如此之慢,因此:

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

其他回答

这就是我在服务器端(如Node.js)添加或更新一些基本url参数时使用的方法。

CoffeScript:

### @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>

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();

以下几点:

合并重复的查询字符串参数 使用绝对和相对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' })

Vianney Bajart的答案是正确的;然而,URL只会工作,如果你有完整的URL端口,主机,路径和查询:

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

URLSearchParams只会在你只传递查询字符串时起作用:

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

如果你有一个不完整或相对的URL,不关心的基础URL,你可以通过?获取查询字符串,然后像这样连接:

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'

Internet Explorer浏览器不兼容。

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。