Re: <help!>EVAL()</help!>

You don't need an eval() here. See http://www.php.net/manual/en/language.variables.variable.php On Mon, 19 Aug 2002, Shaowarrior wrote: > 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 > vdegrand@videotron.ca > > > > -- > PEAR General Mailing List (http://pear.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php >