eval() error

[lentin64]

When I did this PHP Script:
<? if($isput){ $isput=stripSlashes($isput); eval($isput); ?>
I get this error:

Parse error: parse error in /home/stevenc/public_html/stevenchalker/viewphp.php(1) : eval()'d code on line 1

[amcgrath]

For one thing, you don't have a closing } that I can see....

[lentin64]

I have a closing tag.. still an error!

[[deleted]]

It clearly says:
"parse error in ... eval()'d code on line 1
"

So print the content of $isput and find the error.

BTW: this looks very much like you are executing code that comes from a GET or POST operation. I hope you know that that is a very bad idea?

[lentin64]

Well, if you're going to test out the script, you can just go to http://www.qis.net/~ewhitley/stevenchalker/viewphp.php and find the error.

[Anon]

Seeing an example of the eval'd code would help. Plus, if you're using " or ' inside the eval'd code they do need to be escaped... ie the following works

$code="echo \"This is a test\";";
eval($code);

Note that the same rules apply in eval'd code as normal php mode. It is easy to forget the ; at the end of the line in the eval string.

[[deleted]]

I meant that YOU should print the content of the var you are putting into the eval() statement.

Obviously if I test the script that you are having problems with, I wil just see the same error... and I cannot change the sourcecode.

BTW: what you are doing in that script it VERY VERY dangerous!!

[Anon]

I know in ASP eval() is a dynamic way to change a string into a variable. For example user comes, types bob in the input box, clicks submit, now bob is your variable name.

Does it work the same way in php?

[[deleted]]

No. In PHP, whatever you put into the eval() is parsed asif it was PHP code, a bit like the way include() works:

eval('echo "hello";');

eval('$sMyVar = "poo";');
echo $sMyVar;