如何从URL字符串中删除一些参数? [英] How to remove some parameters from an URL string?

问题描述

我有这个var存储一个字符串，该字符串表示一个充满参数的URL.我正在使用AngularJS，但不确定是否有任何有用的模块(或者可能带有普通的JavaScript)可以删除不需要的URL参数，而不必使用正则表达式?

I have this var storing a string that represents a URL full of parameters. I'm using AngularJS, and I'm not sure if there is any useful module (or maybe with plain JavaScript) to remove the unneeded URL parameters without having to use regex?

例如，我需要从以下位置删除&month=05和&year=2017:

For example I need to remove &month=05 and also &year=2017 from:

var url = "at merge ?derivate=21&gear_type__in=13&engine=73&month=05&year=2017"

推荐答案

您可以使用此

You can use this function that take 2 parameters: the param you are trying to remove and your source URL:

function removeParam(key, sourceURL) {
    var rtn = sourceURL.split("?")[0],
        param,
        params_arr = [],
        queryString = (sourceURL.indexOf("?") !== -1) ? sourceURL.split("?")[1] : "";
    if (queryString !== "") {
        params_arr = queryString.split("&");
        for (var i = params_arr.length - 1; i >= 0; i -= 1) {
            param = params_arr[i].split("=")[0];
            if (param === key) {
                params_arr.splice(i, 1);
            }
        }
        rtn = rtn + "?" + params_arr.join("&");
    }
    return rtn;
}

var url = "at merge ?derivate=21&gear_type__in=13&engine=73&month=05&year=2017";

var url2 = removeParam("month", url);
var url3 = removeParam("year", url2);

console.log(url3);

带有正则表达式的替代解决方案

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