Let's say I have the following URL: pic.php?t=1&p=23
When I try phpThumb.php?src=/pic.php?t=1&p=23
, I get Forbidden parameter: p
.
Anyone knows if there's a way around it?
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3 Answers
2
You need to URLEncode that ampersand if you are trying to pass p=23
to pic.php
. Try replacing the &
with %26
.
What you are effectively doing there is passing p=23
as a parameter to phpThumb.php
, which obviously it doesn't like...
What you probably want to do is 'phpThumb.php?src='.urlencode('/pic.php?t=1&p=23');

DaveRandom
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0
You can URL encode the ampersand to make it part of the content of the parameter in the querystring. This way, the whole /pic.php?t=1&p=23
gets passed as the src
.
phpThumb.php?src=/pic.php?t=1%26p=23
The %26
is the encoded version of the &
.
Use urlencode()
.

Pindatjuh
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