AngularJS：避免与$位置URL编码 [英] AngularJS: avoid url encoding with $location

问题描述

我注意到，当我通过一个参数是一个数组$ location.search（），它是连接codeD如下面的例子

I notice that when I passed a parameter that is an array to $location.search(), it was encoded as in the following example

$ location.path（'/ somePath'）搜索（'的ID [] ='[1，2，3]）;

$location.path('/somePath').search('ids[]=', [1, 2, 3]);

变为

/ somePath DS％5B％5D = 1和; DS％5B％5D = 2及DS％5B％5D = 3

/somePath?ds%5B%5D=1&ds%5B%5D=2&ds%5B%5D=3

有没有办法避免URL编码？

Is there a way to avoid the url encoding?

推荐答案

对不起已故的答复，但我打一个类似predicament自己，以为我提供建议。

Excuse the late reply, but I struck a similar predicament myself and thought I'd offer advice.

在总之，如果你打算使用 $ location.search 你无法避免的URL为EN codeD。

In short, if you intend on using $location.search you cannot avoid the URL being encoded.

如果你看看在折角源， location.js 具体而言，你会发现，在函数返回组成的网址（即 LocationHtml5Url ）的所有依靠另一个名为函数调用 toKeyValue ，它将连接code每当他们设置的所有键 - 值对。

If you take a look at the Angular source, location.js specifically, you'll notice that the functions return composed URLs (i.e. LocationHtml5Url) all rely on another function call named toKeyValue, which will encode all key-value pairs whenever they are set.

此外，在本例中使用的情况下，您所提供的，你不需要你的钥匙里的平等。当 $ location.search 被称为有两个参数，它们被视为由角一个键值对。

Furthermore, in the example use case you've provided, you don't need the equals inside your key. When $location.search is called with two parameters, they are treated as a key-value pair by Angular.

如果您需要使用的位置的URL就已经连接codeD之后，你可以随时调用<一个href=\"https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/de$c$cURIComponent\"相对=nofollow> 德codeURIComponent 。

If you need to make use of the location URL after it has been encoded, you could always call decodeURIComponent.

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