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

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 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.branc

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 > L

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 > L