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 Quote Link to comment 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 Quote Link to comment 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. Quote Link to comment Share on other sites More sharing options...
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