Re: FN-FORUM: sql question

date posted 10th August 2003 17:28

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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