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>

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();

你可以使用其中一个:

https://developer.mozilla.org/en-US/docs/Web/API/URL https://developer.mozilla.org/en/docs/Web/API/URLSearchParams

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);

你可以使用url。href或URL . tostring()来获取完整的URL

有时我们看到?在URL结尾，我找到了一些解决方案，生成的结果为file.php?&foo=bar。我想出了我自己的解决方案，以完美地工作!

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

注意:位置。origin不能在IE中工作，这里是它的修复。

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;
        }
    }
});