snorky Posted May 22, 2009 Share Posted May 22, 2009 PHP version = 4.4.9 Pseudo-code for what I'm trying to do: if $path is not a directory then print "" else print $path . "<br />" Actual code attempts: // declare a variable $path = $folder.'/'.$file; What I've tried once the variable is set: if ((isdir($path)=FALSE) if (!isdir($path)) In desperation I tried to see how the test evaluates: $test=var_dump(isdir($path)); echo $test; In each case the result is: Fatal error: Call to undefined function: isdir() in listall.php on line 60 Link to comment https://forums.phpfreaks.com/topic/159314-solved-function-isdir/ Share on other sites More sharing options...
Masna Posted May 22, 2009 Share Posted May 22, 2009 http://us3.php.net/manual/en/function.is-dir.php Link to comment https://forums.phpfreaks.com/topic/159314-solved-function-isdir/#findComment-840222 Share on other sites More sharing options...
snorky Posted May 22, 2009 Author Share Posted May 22, 2009 It helps to spell the function correctly.... is_dir() works better than my original isdir() We can call this solved. Link to comment https://forums.phpfreaks.com/topic/159314-solved-function-isdir/#findComment-840313 Share on other sites More sharing options...
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