根据可变参数值删除部分查询字符串

Question

我正在使用以下代码（在此 post 的帮助下）删除包含 "all"?

的查询字符串参数

转：

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

如何修改代码以使用变量而不是“/all/”？我希望能够轻松地切换出要删除的参数。

现有代码（有效但无变量）：

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();

新代码（无效）

let parametervariable = "/somefoovariable/";
let src = "https://www.foobar.com/page?year=all&language=all&gender=female";
let url = new URL(src);
let re = parametervariable;
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();
 variable):**

新代码报错如下："Uncaught TypeError: re.test is not a function"

Answer 1

你可以使用RegExp构造函数

let parametervariable = "somefoovariable";
let re = new RegExp(parametervariable);

根据可变参数值删除部分查询字符串

Remove Portion of Query String based on a variable parameter value

javascript

regex

query-string