<HELP!>EVAL()</HELP!>
[email protected] ("Shaowarrior") Mon, 19 Aug 2002 18:25:22 -0400
| Newsgroups | php.lang |
|---|---|
| Organization | Kensoft |
| Message-ID | <[email protected]> |
Hi, right now i'm trying to shorten my script and i'm tryin to make this
switch loop :
// Copie le nom du fichier dans un tableau
switch ($ext)
{
case 'jpg':
$imgjpg[$c_jpg] = $repertoire.basename($nomFichier);
$c_jpg++;
break;
case 'gif':
$imggif[$c_gif] = $repertoire.basename($nomFichier);
$c_gif++;
break;
default:
echo ' Le fichier '.$nomFichier.' n\'est pas une image
reconnue.<br>';
break;
}
looks like an only one eval() line of code (separated for your convenience):
eval('$img'."\$ext".'[$c_'."\$ext".'] =
$repertoire.basename($nomFichier);
$c_'."\$ext".'++;
return();');
Well, this don't work! (Guess why i'm here!) PHP debugger says :
PARSE ERROR: parse error, unexcepted T_VARIABLE in c:\[...](20):
eval()'d code on line 1
This is it!
If someone can help me I will appreciate it.
Merci beaucoup!
Shaowarrior
[email protected]