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

重置所有查询字符串

Var params = {params:"val1", params:"val2"}; 让str = jQuery.param(参数); let uri = window.location. reff . 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 " > < /脚本>

其他回答

const params = new URLSearchParams(window.location.search);

params.delete(key)
window.history.replaceState({}, "", decodeURIComponent(`${window.location.pathname}?${params}`));
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;
        }
    }
});

在URL类中有一个内置函数,你可以使用它来轻松处理查询字符串的键/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searchParams函数的更多信息。

IE不支持URL,请检查兼容性

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;
}

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);
      }
    }