Eval

Hey,
Basically, in my code below, I am echoing $code, within $code is:
<img src=\"". $file . "\" />
(stored in the database)

I want to be able to define $file, as I have done earlier in my code. However when I echo out $code, it shows <img src=\"". $file . "\" /> and not <img src=\"WHAT I DEFINE AS $FILE\" />

Someone told me to use eval? How would I do it?

 

$str = '<img src=\"". $file . "\" />';
$file = 'image.jpg';

eval('$new_str = "' . $str . '";');

echo $new_str;

PHP:

 

No! Do not use eval() unless it is really needed. I think this was just a syntax issue.

crazyryan, can you post some more of the code, as I'm sure a non-eval() solution can be found.

 

I tried:

$display = mysql_query("SELECT * FROM `types` WHERE `type` = '" . $type . "'");
$codetype = mysql_fetch_array($display);
$str = $codetype['code'];
eval('$code = "' . $str . '";');
echo $code;

PHP:

But got:

Warning: Unexpected character in input: '\' (ASCII=92) state=1 in /home/phpmedia/public_html/arcade/game.php(28) : eval()'d code on line 1

Parse error: syntax error, unexpected '"' in /home/phpmedia/public_html/arcade/game.php(28) : eval()'d code on line 1

EDIT: This was due to the row in the database, I have fixed it now.

@krt: I really can't see any other way other than doing an if statement for every single file type.

EDIT2: Forgot to say, huge thanks

 

// str_replace() because we only want to escape double quotes.
eval('$code = "' . str_replace('"', '\"', $str) . '";');

PHP:

 

I'm having a bit of trouble with the rows in the database, for one of the codes I have:
<object type=\"application/x-shockwave-flash\" data=\"" . $file . "\" width=\"" . $width ."\" height=\"" . $ height . "\">
<param name=\"movie\" value=\"" . $file . "\" />
</object>

But I'm getting:

Parse error: syntax error, unexpected T_STRING, expecting T_VARIABLE or '$' in /home/phpmedia/public_html/arcade/game.php(28) : eval()'d code on line 1


I'm not using nico's code atm, when I did try the variables weren't being set.

 

=\"" . $ height . "\">

Code (markup):

There's a space between the dollar sign and variable name.

 

Woops, thanks again, nico_swd to the rescue

 

<img src="foldername/<? echo $res['image']?> " >