sylesia Posted December 6, 2006 Share Posted December 6, 2006 Ok, so I am working on code and currently have it in a .html file, and includes some php that accessess a database. Unfortunately I need to really add some fun here with if else statements and such that extend thru the html and incorporate them in. In another words, I need to change my .html to .php but it seems to have issues. I have no problem running it as .html, works fine, accesses the DB, and everything, but when I simply change it to .php, I start to get errors, includingWarning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in C:\Program Files\xampp\htdocs\srh\Junk\cadetyellowedit.php on line 6361 $request = "SELECT $try FROM yellowcard WHERE UserName = '$userName' AND CardNumber = $cardNumber";62 $eval = mysql_query($request, $conn); 63 while($row = mysql_fetch_assoc($eval)){64 $$try = $row[$try]."\n";65 }any ideas why it will not work as .php but does as .html Quote Link to comment Share on other sites More sharing options...
fert Posted December 6, 2006 Share Posted December 6, 2006 that error means your query isn't working Quote Link to comment Share on other sites More sharing options...
marcus Posted December 6, 2006 Share Posted December 6, 2006 [code]$eval = mysql_query($request, $conn) or die(mysql_error());[/code]do you get any errors, try with the die(mysql_error()); Quote Link to comment Share on other sites More sharing options...
sylesia Posted December 6, 2006 Author Share Posted December 6, 2006 The thing I cannot figure out is why its not working. It works fine if the file has a .html extension, works perfectly, but as soon as I change it to .php and no other changes, than it comes with that error. If I add in the or die statement, I getYou have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1Again, that only occurs if it is a .php and not .html extension Quote Link to comment Share on other sites More sharing options...
JasonLewis Posted December 6, 2006 Share Posted December 6, 2006 that is saying that your are checking for a blank variable. try checking $userName has anything and... oh!change:[code=php:0]$request = "SELECT $try FROM yellowcard WHERE UserName = '$userName' AND CardNumber = $cardNumber";[/code]to:[code=php:0]$request = "SELECT $try FROM `yellowcard` WHERE `UserName`= '$userName' AND `CardNumber`='$cardNumber'";[/code]notice how i added the ' and ' around $cardNumber. but also check if both variables contain data. Quote Link to comment Share on other sites More sharing options...
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