Thank you for the response. It looks like that worked! I am going to try it
on the webpage now. I will let everybody know if this works.
Thanks
Rick
PS. to everyone else I want to say thanks. I have run the ones that you have
sent but always came up with an error. If you want me to send you the error
let me know and I will send it to you.
Thanks again
- Original Message -
From: "Shyamal Banerjee" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, November 13, 2002 4:29 AM
Subject: Re: Count Rows in two tables
> If you know the primary keys of the two tables, it may be done as follows:
>
> select count(distinct t1.pk1) + count(distinct t2.pk2) from t1,t2;
> where - pk1 is the primary key column name of table t1 and
> pk2 is the primary key column name of table t2.
>
> SB
> - Original Message -
> From: "Rick Baranowski" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Wednesday, November 13, 2002 8:39 AM
> Subject: Re: Count Rows in two tables
>
>
> > Please forgive me I am new to mySQL, could you give me an example of how
I
> > would do this? The two tables are users and staff
> >
> > Thanks
> > Rick
> >
> >
> > - Original Message -
> > From: "Murad Nayal" <[EMAIL PROTECTED]>
> > Cc: <[EMAIL PROTECTED]>
> > Sent: Tuesday, November 12, 2002 6:58 PM
> > Subject: Re: Count Rows in two tables
> >
> >
> > >
> > > Alan McDonald wrote:
> > > >
> > > > You can't count the join?
> > > > Alan
> > >
> > > if you count the (unqualified) join you'll end up with the product of
> > > the two table counts.
> > >
> > > Murad
> > >
> > > >
> > > > > -Original Message-
> > > > > From: Rick Baranowski [mailto:rickb@;baranconsulting.com]
> > > > > Sent: Wednesday, 13 November 2002 12:10
> > > > > To: [EMAIL PROTECTED]
> > > > > Subject: Count Rows in two tables
> > > > >
> > > > >
> > > > > Hello all,
> > > > >
> > > > > Does anybody have a SQL string to count the rows in two different
> > > > > tables and
> > > > > give you a total number of rows? I have been trying to find an
> > > > > answer for a
> > > > > couple of days and seems like a simple string.
> > > > >
> > > > > Thank you
> > > > >
> > > > > Rick
> > > > >
> > > > >
> > > >
> > -
> > > >
> > >
> > > -
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> >
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