Yes, that's true. But, sub-queries are there for a reason - you can use them
when you need to.
- Original Message -
From: "Mickael" <[EMAIL PROTECTED]>
To: "CF-Talk" <[EMAIL PROTECTED]>
Sent: Monday, December 01, 2003 2:07 PM
Subject: Re: SQL Help on
uot; <[EMAIL PROTECTED]>
Sent: Monday, December 01, 2003 12:37 PM
Subject: Re: SQL Help on Join
> Mickael writes:
> > Select Count(*) as TotalAccounts, Sum(Orig_amt) as SumOrig_AMT,
Sum(Cur_Bal) as SumCurBal, Sum(Pmt_amt) as SUMPmtamt
> > from Client_debt
&
t;
Sent: Monday, December 01, 2003 12:37 PM
Subject: Re: SQL Help on Join
> Mickael writes:
> > Select Count(*) as TotalAccounts, Sum(Orig_amt) as SumOrig_AMT,
Sum(Cur_Bal) as SumCurBal, Sum(Pmt_amt) as SUMPmtamt
> > from Client_debt
> > Left outer join Client_pmt on client_d
Typically you don't count *
But you would do it this way... count(tablename.*)
-Novak
-Original Message-
From: Mickael [mailto:[EMAIL PROTECTED]
Sent: Monday, December 01, 2003 10:18 AM
To: CF-Talk
Subject: Re: SQL Help on Join
how do I specify the tablename with the
Hey Scott,
I am still getting the two records for an account, if there are the records in the client_pmt table, How does distinct work?
- Original Message -
From: Scott Weikert
To: CF-Talk
Sent: Monday, December 01, 2003 1:37 PM
Subject: Re: SQL Help on Join
Mickael
how do I specify the tablename with the Count(*)?
- Original Message -
From: Scott Weikert
To: CF-Talk
Sent: Monday, December 01, 2003 1:37 PM
Subject: Re: SQL Help on Join
Mickael writes:
> Select Count(*) as TotalAccounts, Sum(Orig_amt) as SumOrig_AMT, Sum(Cur_Bal)
Mickael writes:
> Select Count(*) as TotalAccounts, Sum(Orig_amt) as SumOrig_AMT, Sum(Cur_Bal) as SumCurBal, Sum(Pmt_amt) as SUMPmtamt
> from Client_debt
> Left outer join Client_pmt on client_debt.debt_id = client_pmt.debt_id
You're specifying the table names of the fields in your JOIN but not in
Hello All,
I am not getting the results that I am looking for may query.
Here is my sql
Select Count(*) as TotalAccounts, Sum(Orig_amt) as SumOrig_AMT, Sum(Cur_Bal) as SumCurBal, Sum(Pmt_amt) as SUMPmtamt
from Client_debt
Left outer join Client_pmt on client_debt.debt_id = client_pmt.debt_id
Th
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