CodingForums.com - View Single Post - Not displaying the first item in an array

Fou-Lu · 09-13-2012, 05:57 PM

PHP Code:


			
            $info2 = mysql_fetch_array($data2);
// . . .
            while($info2 = mysql_fetch_array($data2)){

You've already fetched the first row into $info2. Calling it in the while now starts at the second record.
If you need to keep this if branch check (invert it with a ! and remove the empty braces instead of using an else), then change the while loop into a do/while loop and it will function properly. Alternatively use mysql_data_seek to change your record. Personally I'd alter the query since you don't need to use that if/else branch there at all; simply add a check on that field for either not an empty string (if its an empty string), or not is null check if its null.

09-13-2012, 05:57 PM	PM User \| #3
Fou-Lu God Emperor Join Date: Sep 2002 Location: Saskatoon, Saskatchewan Posts: 15,635 Thanks: 4 Thanked 2,448 Times in 2,417 Posts	PHP Code: `$info2 = mysql_fetch_array($data2); // . . . while($info2 = mysql_fetch_array($data2)){` You've already fetched the first row into $info2. Calling it in the while now starts at the second record. If you need to keep this if branch check (invert it with a ! and remove the empty braces instead of using an else), then change the while loop into a do/while loop and it will function properly. Alternatively use mysql_data_seek to change your record. Personally I'd alter the query since you don't need to use that if/else branch there at all; simply add a check on that field for either not an empty string (if its an empty string), or not is null check if its null.