RE: [PHP-DB] Putting a string into a variable.

2003-03-28 Thread Jennifer Goodie
The part where $i is never equal to 1 might be the problem. You might want to try incrementing your sentinel when using loops. -Original Message- From: Info_Best-IT [mailto:[EMAIL PROTECTED] Sent: Friday, March 28, 2003 11:13 AM To: [EMAIL PROTECTED] Subject: [PHP-DB] Putting a string

RE: [PHP-DB] Putting a string into a variable.

2003-03-28 Thread Hutchins, Richard
You might be getting an empty result because you don't have anything in $stringarray[1]. You set $i = 0 then have a while loop that does something while $i 6. But you never go through and increment $i at all so it's always equal to 0. You might get some type of result if you echoed

RE: [PHP-DB] Putting a string into a variable.

2003-03-28 Thread Hutchins, Richard
- From: Hutchins, Richard [mailto:[EMAIL PROTECTED] Sent: Friday, March 28, 2003 2:22 PM To: 'Info_Best-IT'; [EMAIL PROTECTED] Subject: RE: [PHP-DB] Putting a string into a variable. You might be getting an empty result because you don't have anything in $stringarray[1]. You set $i = 0

Re: [PHP-DB] Putting a string into a variable.

2003-03-28 Thread Ronan Chilvers
Hi Tim Comments inline... $stringarray[] = array(); $i = 0; while ( $i 6 ){ $stringarray[$i] = table cellspacing=\0\trtd.$this-differentarray[0] [0]./td/tr/table; } When I echo $stringarray[1] it comes up empty... Is there a special trick to get a You're not incrementing $i.