Subject: Re: SQL Help - Answered
I'm going to have to look at something. I'm still not getting what I
anticipated.
There are over 3700 records of which 1775 of them are distinct values
for RATE. However, I am only getting values of 1 for COUNT(DISTINCT
rate) as rateCount. I
Thank you
Greg Morphis wrote:
You need a group by in your query..
SELECT count(DISTINCT rate) as rateCount, rate
FROMmyrates
WHERE my_code = 385 and year = 2005
GROUP BY rate
ORDER BY rate
On 4/3/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Why do the
No problem, if you want to know why take a look at aggregate
functions, which is what count is, as well as others..
On 4/3/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Thank you
Greg Morphis wrote:
You need a group by in your query..
SELECT count(DISTINCT rate) as rateCount, rate
I'm going to have to look at something. I'm still not getting what I
anticipated.
There are over 3700 records of which 1775 of them are distinct values
for RATE. However, I am only getting values of 1 for COUNT(DISTINCT
rate) as rateCount. I was trying to find out how many records are
can you provide a sample of your data and the way your table is
designed (column name, type)?
Thanks
On 4/3/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
I'm going to have to look at something. I'm still not getting what I
anticipated.
There are over 3700 records of which 1775 of them are
that and see if you're any closer to what you want.
-- Josh
- Original Message -
From: [EMAIL PROTECTED] [EMAIL PROTECTED]
To: CF-Talk cf-talk@houseoffusion.com
Sent: Tuesday, April 03, 2007 2:09 PM
Subject: Re: SQL Help - Answered
I'm going to have to look at something. I'm still
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