根据JS值，使用JS删除部分查询字符串，而无需重新加载页面 [英] Remove Portion of Query String using JS based on their value without page reload

问题描述

如何仅根据参数从URL中删除某些参数? 例如，如果我想以编程方式删除所有包含值"all"的参数集，即当Ajax事件完成时.

How can I remove just certain parameters from an URL based on their parameters? For example, If I want to programmatically remove any parameter sets that contain the value "all" i.e. when an Ajax event completes.

www.foobar.com/page?year=all&language=all&gender=女性 到: www.foobar.com/page?gender=female

www.foobar.com/page?year=all&language=all&gender=female to: www.foobar.com/page?gender=female

使用JS或jQuery.

Using JS or jQuery.

推荐答案

您可以使用URL()构造函数URLSearchParams()来获取URL的查询字符串，作为与查询字符串参数相对应的属性和值数组的Array ，Array.prototype.filter()和RegExp.prototype.test()从数组中删除查询字符串参数，然后再次URLSearchParams()重建已过滤的查询字符串参数

You can use URL() constructor, URLSearchParams() to get query string of URL as an Array of arrays of properties and values corresponding to query string parameters, Array.prototype.filter() and RegExp.prototype.test() to remove query string parameters from array, and URLSearchParams() again to reconstruct filtered query string parameters

let src = "https://www.foobar.com/page?year=all&language=all&gender=female";
let url = new URL(src);
let re = /all/;
let props = [...new URLSearchParams(url.search)]
            .filter(([key, prop]) => !re.test(prop));
url = url.origin + url.pathname;
let params = new URLSearchParams();
props.forEach(([key, prop]) => params.set(key, prop));
url += "?" + params.toString();

console.log(url);

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