[PHP] joins issues again

2008-04-08 Thread Steven Macintyre
Hi all, I have the following SQL statement; SELECT count( salesID ) AS count, branch_name, company_name, branch.branchID FROM sales LEFT JOIN IGuser ON sales.IGuid = IGuser.IGuid LEFT JOIN branch ON IGuser.branchID = branch.branchID LEFT JOIN company ON

Re: [PHP] joins issues again

2008-04-08 Thread Andrew Ballard
On Tue, Apr 8, 2008 at 7:28 AM, Steven Macintyre [EMAIL PROTECTED] wrote: Hi all, I have the following SQL statement; SELECT count( salesID ) AS count, branch_name, company_name, branch.branchID FROM sales LEFT JOIN IGuser ON sales.IGuid = IGuser.IGuid LEFT JOIN

Re: [PHP] joins issues again

2008-04-08 Thread Daniel Brown
On Tue, Apr 8, 2008 at 7:28 AM, Steven Macintyre [EMAIL PROTECTED] wrote: Hi all, I have the following SQL statement; SELECT count( salesID ) AS count, branch_name, company_name, branch.branchID FROM sales LEFT JOIN IGuser ON sales.IGuid = IGuser.IGuid LEFT JOIN

Re: [PHP] joins issues again

2008-04-08 Thread Wolf
Steven Macintyre [EMAIL PROTECTED] wrote: Hi all, I have the following SQL statement; SELECT count( salesID ) AS count, branch_name, company_name, branch.branchID FROM sales LEFT JOIN IGuser ON sales.IGuid = IGuser.IGuid LEFT JOIN branch ON IGuser.branchID =

Re: [PHP] joins issues again

2008-04-08 Thread Mark J. Reed
On Tue, Apr 8, 2008 at 7:28 AM, Steven Macintyre [EMAIL PROTECTED] wrote: I have the following SQL statement; ... and this relates to PHP how? SELECT count( salesID ) AS count, branch_name, company_name, branch.branchID That doesn't make sense. You're selecting a group function (COUNT)

Re: [PHP] joins issues again

2008-04-08 Thread Jim Lucas
Steven Macintyre wrote: Hi all, I have the following SQL statement; SELECT count( salesID ) AS count, branch_name, company_name, branch.branchID FROM sales LEFT JOIN IGuser ON sales.IGuid = IGuser.IGuid LEFT JOIN branch ON IGuser.branchID = branch.branchID LEFT