如何在角度js中获取url参数

Here's my url

这是我的网址

localhost/project/#/showprofile/18

I want to display the parameter 18 in my view

我想在视图中显示参数18

In the app.js i have

在app.js我有

.when('/showprofile/:UserID', {
     title: 'Show User Profile',
     templateUrl: 'views/layout/showprofile.php',
     controller: 'authCtrl',
     })

Here the page showprofile.php is displaying, but suddenly the url goes like this

在这里显示showprofile.php页面,但突然网址就像这样

localhost/project/#/showprofile/:UserId

How can i get the value 18 inside the showprofile.php and make the url as it is i.e.,

如何在showprofile.php中获取值18并使其成为原样,即,

localhost/project/#/showprofile/18

3 个解决方案

#1

Use $routeParams to get the parameter value:

使用$ routeParams获取参数值:

$routeParams.UserID

Make sure you inject $routeParams before using it.

确保在使用之前注入$ routeParams。

EDIT How can i get the value 18 inside the showprofile.php

编辑如何在showprofile.php中获取值18

<?php
    $link = $_SERVER['PHP_SELF']; // Get current URL
    $link_array = explode('/', $link); // Split by /
    echo $page = end($link_array); // Get last element from array

?>

3 个解决方案

#1

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