Re: FN-FORUM: sql question

date posted 10th August 2003 17:48

Also if you are using MySQL you can't use nested queries (unless you're
using the alpha version of 4.1.0 and the syntax is different anyway).
And most other database do not perform automatic string to integer
conversion. So you may still have a problem unless project_id is a
string.

On Sun, 2003-08-10 at 17:56, PAMELA WHITTAKER wrote:
> OOPS - I have just realised that "SELECT Employee_Id FROM
> greensplash_employees_per_project where Project_Id = '18' " won't find
> anything. No wonder it is failing.
>
> Thanks
> Pam
>
>
>
> ----- Original Message -----
> From: "Alan Sheppard" [EMAIL REMOVED]
> To: [EMAIL REMOVED]
> Sent: Sunday, August 10, 2003 5:49 PM
> Subject: RE: FN-FORUM: sql question
>
>
> >
> > Pam
> >
> > Is project_id a numeric and your testing against a string '18' ?
> >
> > Alan S
> >
> > On 10 August 2003 17:42, PAMELA WHITTAKER [EMAIL REMOVED]
> wrote:
> > >
> > > I am getting the following error
> > > You have an error in your SQL syntax near 'SELECT Employee_Id FROM
> > > greensplash_employees_per_project where Project_Id = '18' at line 1.
> > >
> > > I have set the sql string as follows
> > >
> > > $sqlstring = "SELECT Employee_Id, Employee_Forename, Employee_Surname
> FROM
> > > greensplash_employees_details where Employee_Status = 'Active' AND
> > > Employee_Id NOT IN (SELECT Employee_Id FROM
> > > greensplash_employees_per_project where Project_Id = '" .
> > > $row_Project_Id['Project_Id'] . "') Order By Employee_Forename";
> > >
> > > and echo prints out
> > >
> > > SELECT Employee_Id, Employee_Forename, Employee_Surname FROM
> > > greensplash_employees_details where Employee_Status = 'Active' AND
> > > Employee_Id NOT IN (SELECT Employee_Id FROM
> > > greensplash_employees_per_project where Project_Id = '18') Order By
> > > Employee_Forename
> > >
> >
> >
> >
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