RE: [PHP] Returning a variable from a function problem

> -----Original Message----- > From: ODCS [mailto:odcs@rogers.com] > Sent: 02 April 2003 08:10 > > I have a form where the user inputs information - the code > below is the > error checking for one of the fields. The first IF statement > just checks > that the filed is not empty and works fine. > > Then I check a function I made to see if the user is already > in a mysql > database and the function returns 0, 1, or 2 depending on the various > conditions. The first elseif works as I would expect - but it > seems that the > next two elseif's are just being ignored. > > Could someone possibly explain what I have wrong here. Thanks > > if (empty($form["name"])) { $error["name"] = "*"; > $message["name"] = > "Required field!"; > } > elseif($duplicate = duplicate_name($form["name"]) == 0) { This statement assigns the result of the comparison duplicate_name($form["name"]) == 0 to $duplicate (giving FALSE if this branch isn't executed, and so never satisfying the tests for 1 or 2 below). I suspect that what you mean is: elseif(($duplicate = duplicate_name($form["name"])) == 0) { > $error["name"] = "*"; $message["name"] = "Screen name > already used!"; > . > } > elseif ($duplicate == 1) { > $error["name"] = "*"; $message["name"] = "Database > error > - Please try > again in a few moments!"; > } > elseif ($duplicate == 2) { > $error["name"] = "*"; $message["name"] = "Server error - > > Please try > again in a few moments!"; > } Cheers! Mike --------------------------------------------------------------------- Mike Ford, Electronic Information Services Adviser, Learning Support Services, Learning & Information Services, JG125, James Graham Building, Leeds Metropolitan University, Beckett Park, LEEDS, LS6 3QS, United Kingdom Email: m.ford@lmu.ac.uk Tel: +44 113 283 2600 extn 4730 Fax: +44 113 283 3211