note 41508 deleted from function.gettype by nlopess

From: Date: Tue, 13 Apr 2004 10:46:44 +0000
Subject: note 41508 deleted from function.gettype by nlopess
References: 1  Groups: php.notes 
Request: Send a blank email to php-notes+get-68060@lists.php.net to get a copy of this message
Note Submitter: MC_Breit @phpws.org ---- Here is an alternative varInformation function, I use it to debug my scripts. <?php /****** * varInfo debug function by MC_Breit * from phpws.org * ! Use varInfo('$var') instead of varInfo($var) ! */ //Define some strings to test the script $teststring = 'test'; $testinteger = 123; $testarray = array('Test1', 'Test2'); //A String varInfo('$teststring'); echo'<br>'; //A Integer varInfo('$testinteger'); echo'<br>'; //A Array varInfo('$testarray'); echo'<br>'; //A Empty var.. varInfo('$empty'); echo'<br>'; //Producing an error ;) if( !varInfo('A name for a variable, which is not valid.. $)(') ) echo varInfo('ERROR'); function varInfo($vtitle) { static $error; if( $vtitle == 'ERROR' ) { return $error; } //Check validation of given variable name if( !preg_match('/^\$[a-zA-Z0-9]+$/', $vtitle) ) { $error = '<b>Syntax Error:</b> "'.$vtitle.'" is not a valid name for a var!'; return FALSE; } //Get var values: eval('global '.$vtitle.'; $vvalue = '.$vtitle.';'); //Get vartype if( is_array($vvalue) ) { $vtype = 'Array'; } elseif( empty($vvalue) ) { print 'The var "'.$vtitle.'" is empty or doesnt exist.'."\n"; return TRUE; } elseif ( is_string($vvalue) ) { $vtype = 'String'; } elseif ( is_integer($vvalue) ) { $vtype = 'Integer'; } elseif ( is_resource($vvalue) ) { $vtype = 'Resource'; } elseif ( is_object($vvalue) ) { $vtype = 'Object'; } elseif ( is_null($vvalue) ) { print 'The var "'.$vtitle.'" is NULL.'."\n"; return TRUE; } else { $error = '<b>General Error:</b> PHP says the variable '.$vartitle.' is "'.gettype($varvalue).'" but can\'t proof this!'; return FALSE; } //Print the result: print 'The var "'.$vtitle.'" is a '.$vtype.' and its value is: '; eval(($vtype == 'Array') ? 'print_r($vvalue);' : 'print $vvalue."\n";'); return TRUE; } ?> Returns: The var "$teststring" is a String and its value is: test The var "$testinteger" is a Integer and its value is: 123 The var "$testarray" is a Array and its value is: Array ( [0] => Test1 [1] => Test2 ) The var "$empty" is empty or doesnt exist. Syntax Error"A name for a variable, which is not valid.. $)(" is not a valid name for a var!

« previous php.notes (#68060) next »