问题描述
关于$ location.search的说，
Regarding $location.search, the docs say,
返回搜索部分（如对象）当前URL时不带任何参数调用。
在我的网址，我的查询字符串有一个参数？test_user_bLzgB'无值。也'$ location.search（）返回一个对象。我如何获得的实际文本？
In my url, my query string has a param '?test_user_bLzgB' without a value. Also '$location.search()' returns an object. How do I get the actual text?
推荐答案
不知道是否已经改变，因为公认的答案被接受，但却是可能的。
Not sure if it has changed since the accepted answer was accepted, but it is possible.
$ location.search（）将返回键 - 值对，同样对作为查询字符串的对象。没有值的一个关键是刚刚存储在对象为真。在这种情况下，对象将是：
$location.search() will return an object of key-value pairs, the same pairs as the query string. A key that has no value is just stored in the object as true. In this case, the object would be:
{"test_user_bLzgB": true}
您可以用 $ location.search（）。test_user_bLzgB
示例（具有较大的查询字符串）：<一href=\"http://fiddle.jshell.net/TheSharpieOne/yHv2p/4/show/?test_user_bLzgB&somethingElse&also&something=Somethingelse\">http://fiddle.jshell.net/TheSharpieOne/yHv2p/4/show/?test_user_bLzgB&somethingElse&also&something=Somethingelse
Example (with larger query string): http://fiddle.jshell.net/TheSharpieOne/yHv2p/4/show/?test_user_bLzgB&somethingElse&also&something=Somethingelse
注：由于哈希（因为它会去
As pointed out in the comments by @Naresh and @DavidTchepak, the $locationProvider also needs to be configured properly: https://code.angularjs.org/1.2.23/docs/guide/$location#-location-service-configuration
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