Re: [PHP] Resource ID# 5 and Resource id #5 8561 errors....help

2008-12-12 Thread Jim Lucas
Terion Miller wrote: > On Fri, Dec 12, 2008 at 5:08 PM, Daniel P. Brown > wrote: > >> On Fri, Dec 12, 2008 at 18:03, Terion Miller >> wrote: >>> Well I did some changes and I must be learning because although I have >> the >>> same error I don't have new ones... >>> so now the code is like this:

Re: [PHP] Resource ID# 5 and Resource id #5 8561 errors....help

2008-12-12 Thread Terion Miller
On Fri, Dec 12, 2008 at 5:08 PM, Daniel P. Brown wrote: > On Fri, Dec 12, 2008 at 18:03, Terion Miller > wrote: > > > > Well I did some changes and I must be learning because although I have > the > > same error I don't have new ones... > > so now the code is like this: > > $sql = "SELECT * FROM

Re: [PHP] Resource ID# 5 and Resource id #5 8561 errors....help

2008-12-12 Thread Daniel P. Brown
On Fri, Dec 12, 2008 at 18:03, Terion Miller wrote: > > Well I did some changes and I must be learning because although I have the > same error I don't have new ones... > so now the code is like this: > $sql = "SELECT * FROM `importimages` WHERE `Category` = 'Obits'"; > $result = mysql_query($sql)

Re: [PHP] Resource ID# 5 and Resource id #5 8561 errors....help

2008-12-12 Thread Terion Miller
On Fri, Dec 12, 2008 at 4:52 PM, Terion Miller wrote: > > > On Fri, Dec 12, 2008 at 4:02 PM, Daniel P. Brown < > daniel.br...@parasane.net> wrote: > >> On Fri, Dec 12, 2008 at 16:54, Terion Miller >> wrote: >> > >> > $query = "SELECT * FROM importimages WHERE Category='Obits' "; >> > $result = my

Re: [PHP] Resource ID# 5 and Resource id #5 8561 errors....help

2008-12-12 Thread Terion Miller
On Fri, Dec 12, 2008 at 4:02 PM, Daniel P. Brown wrote: > On Fri, Dec 12, 2008 at 16:54, Terion Miller > wrote: > > > > $query = "SELECT * FROM importimages WHERE Category='Obits' "; > > $result = mysql_query ($query); > > > > $arr = mysql_fetch_row($result); > > $result2 = $arr[0]; > > echo ($re

Re: [PHP] Resource ID# 5 and Resource id #5 8561 errors....help

2008-12-12 Thread Daniel P. Brown
On Fri, Dec 12, 2008 at 16:54, Terion Miller wrote: > > $query = "SELECT * FROM importimages WHERE Category='Obits' "; > $result = mysql_query ($query); > > $arr = mysql_fetch_row($result); > $result2 = $arr[0]; > echo ($result2); Try this to get yourself started: $v) { echo stripsl

[PHP] Resource ID# 5 and Resource id #5 8561 errors....help

2008-12-12 Thread Terion Miller
Hey there, okay I ran into these and need some tips, pointers etc... First I was getting the Resource ID#5 error with this query: $query="SELECT * FROM importimages WHERE Category='Obits'"; $result = mysql_query($query); so then I read how mysql_query returns a resource, so I tried this: $query