用javascript我怎么能添加一个查询字符串参数的url,如果不存在或如果它存在,更新当前值?我使用jquery为我的客户端开发。


当前回答

给出Gal和tradyblix建议的修改window.location.search的代码示例:

var qs = window.location.search || "?";
var param = key + "=" + value; // remember to URI encode your parameters
if (qs.length > 1) {
    // more than just the question mark, so append with ampersand
    qs = qs + "&";
}
qs = qs + param;
window.location.search = qs;

其他回答

通过使用jQuery,我们可以做到如下所示

var query_object = $.query_string;
query_object["KEY"] = "VALUE";
var new_url = window.location.pathname + '?'+$.param(query_object)

在变量new_url中我们将有新的查询参数。

参考:http://api.jquery.com/jquery.param/

是的,我有一个问题,我的查询字符串会溢出和复制,但这是由于我自己的懒惰。所以我玩了一点,并研究了一些js jquery(实际上是sizzle)和c#魔法。

所以我才意识到,在服务器已经完成了传递的值,这些值不再重要,没有重用,如果客户端想做同样的事情,显然它将总是一个新的请求,即使它是相同的参数被传递。这些都是客户端,所以一些缓存/cookie等在这方面可能很酷。

JS:

$(document).ready(function () {
            $('#ser').click(function () {
                SerializeIT();
            });
            function SerializeIT() {
                var baseUrl = "";
                baseUrl = getBaseUrlFromBrowserUrl(window.location.toString());
                var myQueryString = "";
                funkyMethodChangingStuff(); //whatever else before serializing and creating the querystring
                myQueryString = $('#fr2').serialize();
                window.location.replace(baseUrl + "?" + myQueryString);
            }
            function getBaseUrlFromBrowserUrl(szurl) {
                return szurl.split("?")[0];
            } 
            function funkyMethodChangingStuff(){
               //do stuff to whatever is in fr2
            }
        });

HTML:

<div id="fr2">
   <input type="text" name="qURL" value="http://somewhere.com" />
   <input type="text" name="qSPart" value="someSearchPattern" />
</div>
<button id="ser">Serialize! and go play with the server.</button>

C#:

    using System.Web;
    using System.Text;
    using System.Collections.Specialized;

    public partial class SomeCoolWebApp : System.Web.UI.Page
    {
        string weburl = string.Empty;
        string partName = string.Empty;

        protected void Page_Load(object sender, EventArgs e)
        {
            string loadurl = HttpContext.Current.Request.RawUrl;
            string querySZ = null;
            int isQuery = loadurl.IndexOf('?');
            if (isQuery == -1) { 
                //If There Was no Query
            }
            else if (isQuery >= 1) {
                querySZ = (isQuery < loadurl.Length - 1) ? loadurl.Substring(isQuery + 1) : string.Empty;
                string[] getSingleQuery = querySZ.Split('?');
                querySZ = getSingleQuery[0];

                NameValueCollection qs = null;
                qs = HttpUtility.ParseQueryString(querySZ);

                weburl = qs["qURL"];
                partName = qs["qSPart"];
                //call some great method thisPageRocks(weburl,partName); or whatever.
          }
      }
  }

好的,批评是受欢迎的(这是一个夜间调制,所以请随时注意调整)。如果这有帮助的话,竖起大拇指,快乐编码。

没有重复,每个请求都是唯一的,因为你修改了它,并且由于它的结构,很容易从dom内动态添加更多的查询。

我知道这是相当旧的,但我想把我的工作版本在这里。

function addOrUpdateUrlParam(uri, paramKey, paramVal) { var re = new RegExp("([?&])" + paramKey + "=[^&#]*", "i"); if (re.test(uri)) { uri = uri.replace(re, '$1' + paramKey + "=" + paramVal); } else { var separator = /\?/.test(uri) ? "&" : "?"; uri = uri + separator + paramKey + "=" + paramVal; } return uri; } jQuery(document).ready(function($) { $('#paramKey,#paramValue').on('change', function() { if ($('#paramKey').val() != "" && $('#paramValue').val() != "") { $('#uri').val(addOrUpdateUrlParam($('#uri').val(), $('#paramKey').val(), $('#paramValue').val())); } }); }); <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <input style="width:100%" type="text" id="uri" value="http://www.example.com/text.php"> <label style="display:block;">paramKey <input type="text" id="paramKey"> </label> <label style="display:block;">paramValue <input type="text" id="paramValue"> </label>

说明修改后的@elreimundo

感谢现代javascript, node.js和浏览器的支持,我们可以摆脱第三方库的漩涡(jquery, query-string等)和DRY自己。

下面是javascript(node.js)和typescript版本的函数,用于添加或更新给定url的查询参数:

Javascript

const getUriWithParam = (baseUrl, params) => { const Url = new URL(baseUrl); const urlParams = new URLSearchParams(Url.search); for (const key in params) { if (params[key] !== undefined) { urlParams.set(key, params[key]); } } Url.search = urlParams.toString(); return Url.toString(); }; console.info('expected: https://example.com/?foo=bar'); console.log(getUriWithParam("https://example.com", {foo: "bar"})); console.info('expected: https://example.com/slug?foo=bar#hash'); console.log(getUriWithParam("https://example.com/slug#hash", {foo: "bar"})); console.info('expected: https://example.com/?bar=baz&foo=bar'); console.log(getUriWithParam("https://example.com?bar=baz", {foo: "bar"})); console.info('expected: https://example.com/?foo=baz&bar=baz'); console.log(getUriWithParam("https://example.com?foo=bar&bar=baz", {foo: "baz"}));

打印稿


const getUriWithParam = (
  baseUrl: string,
  params: Record<string, any>
): string => {
  const Url = new URL(baseUrl);
  const urlParams: URLSearchParams = new URLSearchParams(Url.search);
  for (const key in params) {
    if (params[key] !== undefined) {
      urlParams.set(key, params[key]);
    }
  }
  Url.search = urlParams.toString();
  return Url.toString();
};

对于React Native

URL在React Native中没有实现。所以你必须事先安装react-native-url-polyfill。

对于对象参数

请看这个答案中的第二个解决方案

下面是使用锚HTML元素的内置属性的另一种方法:

处理多值参数。 没有修改#片段或查询字符串本身以外的任何内容的风险。 可能会更容易读?但是它更长。

var a = document.createElement('a'), getHrefWithUpdatedQueryString = function(param, value) { return updatedQueryString(window.location.href, param, value); }, updatedQueryString = function(url, param, value) { /* A function which modifies the query string by setting one parameter to a single value. Any other instances of parameter will be removed/replaced. */ var fragment = encodeURIComponent(param) + '=' + encodeURIComponent(value); a.href = url; if (a.search.length === 0) { a.search = '?' + fragment; } else { var didReplace = false, // Remove leading '?' parts = a.search.substring(1) // Break into pieces .split('&'), reassemble = [], len = parts.length; for (var i = 0; i < len; i++) { var pieces = parts[i].split('='); if (pieces[0] === param) { if (!didReplace) { reassemble.push('&' + fragment); didReplace = true; } } else { reassemble.push(parts[i]); } } if (!didReplace) { reassemble.push('&' + fragment); } a.search = reassemble.join('&'); } return a.href; };