URL contain % ,%2f would be a / and the %3d a = In yii?

Question

My url in yii is:

http://localhost/php_pro_106/reload/ByCustomer/mJYwIzoaIGe0R8lAVCqPhG%2Fg0jJFWjiWWdPnkq5VDlY%3D

what should I have to do in url manager to get this,My current rule is:

'<controller:\w+>/ByCustomer/<giftcode>'=>'<controller>/ByCustomer',

Actually problem is that I get :

The requested URL     /php_pro_106/reload/ByCustomer/mJYwIzoaIGe0R8lAVCqPhG/g0jJFWjiWWdPnkq5VDlY= was     not found on this server.

It is due to %3D %2F in the URL.What I have to do to make it work fine?

@Blubberguy22 I dont know .But it is encrypt form of data. – unknownbits Aug 11 '14 at 12:56 — unknownbits, Aug 11 '14 at 12:56

Marek · Accepted Answer · 2014-08-11T13:57:17.430

0

I know apache has problem with %2F in URL, it will always reply with 404, php script is not executed. I never found a way around it. But that was some years ago.

EDIT:

There's a solution, but it requires you can edit apache configuration file for your virtual host at least (I didn't have that at that time). Then you have to add AllowEncodedSlashes On inside your <VirtualHost>. Or you can also set it globaly.

edited Aug 11 '14 at 13:57

answered Aug 11 '14 at 13:07

Marek

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I did not use this , I have posted my solution. – unknownbits Aug 11 '14 at 17:33

score 0 · Answer 2 · edited May 23 '17 at 10:33

I got the solution of it,For that I have to change my encode and decode method:

function base64_url_encode($input) {
 return strtr(base64_encode($input), '+/=', '-_,');
}

function base64_url_decode($input) {
 return base64_decode(strtr($input, '-_,', '+/='));
}

The problem was due to some characters which are there in encoded Url.So, I have to replace these characters.

These links Help me:

Stackoverflow

and

google Group

URL contain % ,%2f would be a / and the %3d a = In yii?

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