In article [EMAIL PROTECTED],
Phil Rhoades [EMAIL PROTECTED] writes:
People,
select count(*) as cnt, name from tst group by name having count(*) = 1
This worked for my basic example but not for my actual problem - I get
column comment must appear in the GROUP BY clause or be used in an
Harald Fuchs [EMAIL PROTECTED] writes:
If you want to select both columns, but have uniqueness over the first
only, you can use a derived table:
SELECT tbl.name, tbl.comment
FROM tbl
JOIN (SELECT name FROM tbl GROUP BY name HAVING count(*) = 1) AS t
ON t.name = tbl.name
Or use the
Hello
try
SELECT DISTINCT col FROM table
Pavel
On 27/01/2008, Phil Rhoades [EMAIL PROTECTED] wrote:
People,
I want to select from a table ONLY unique records ie if a column has
values:
1
2
3
3
4
5
I want ONLY these records returned:
1
2
4
5
Thanks,
Phil.
--
Philip
People,
I want to select from a table ONLY unique records ie if a column has
values:
1
2
3
3
4
5
I want ONLY these records returned:
1
2
4
5
Thanks,
Phil.
--
Philip Rhoades
Pricom Pty Limited (ACN 003 252 275 ABN 91 003 252 275)
GPO Box 3411
Sydney NSW 2001
Australia
Fax:
Phil Rhoades wrote:
People,
I want to select from a table ONLY unique records ie if a column has
values:
1
2
3
3
4
5
I want ONLY these records returned:
1
2
4
5
SELECT count(*) as cnt,a,b,c FORM yourtable
GROUP BY a,b,c
HAVING cnt=1
should do.
Regards
Tino
Pavel,
You didn't read my note properly - your query gives:
1
2
3
4
5
I want:
1
2
4
5
Phil.
On Sun, 2008-01-27 at 15:10 +0100, Pavel Stehule wrote:
Hello
try
SELECT DISTINCT col FROM table
Pavel
On 27/01/2008, Phil Rhoades [EMAIL PROTECTED] wrote:
People,
I want to select
Tino,
On Sun, 2008-01-27 at 15:16 +0100, Tino Wildenhain wrote:
Phil Rhoades wrote:
People,
I want to select from a table ONLY unique records ie if a column has
values:
1
2
3
3
4
5
I want ONLY these records returned:
1
2
4
5
SELECT count(*) as
On 27/01/2008, Phil Rhoades [EMAIL PROTECTED] wrote:
Tino,
On Sun, 2008-01-27 at 15:16 +0100, Tino Wildenhain wrote:
Phil Rhoades wrote:
People,
I want to select from a table ONLY unique records ie if a column has
values:
1
2
3
3
4
5
I want ONLY these
Guys,
On Sun, 2008-01-27 at 17:38 +0100, Pavel Stehule wrote:
On 27/01/2008, Phil Rhoades [EMAIL PROTECTED] wrote:
Tino,
On Sun, 2008-01-27 at 15:16 +0100, Tino Wildenhain wrote:
Phil Rhoades wrote:
People,
I want to select from a table ONLY unique records ie if a column
On Mon, Jan 28, 2008 at 03:32:18AM +1100, Phil Rhoades wrote:
SELECT count(*) as cnt, name FRoM tst GROUP BY name HAVING cnt = 1 ;
ERROR: column cnt does not exist
LINE 1: ...ount(*) as cnt, name FRoM tst GROUP BY name HAVING cnt = 1 ;
having count(*) = 1;
depesz
--
quicksil1er: postgres
People,
select count(*) as cnt, name from tst group by name having count(*) = 1
This worked for my basic example but not for my actual problem - I get
column comment must appear in the GROUP BY clause or be used in an
aggregate function errors so I have a related question:
With table:
name
Hi Phil,
Each of columns that you specify in your SELECT clause, must also
appear in the GROPU BY clause.
SELECT COUNT(*) AS cnt, name, comment, ...
FROM tst
GROUP BY name, comment, ...
HAVING COUNT(*) = 1;
Phil Rhoades wrote:
People,
select count(*) as cnt, name from tst group by
On Sunday 27 January 2008 10:56:18 am Mike Ginsburg wrote:
Hi Phil,
Each of columns that you specify in your SELECT clause, must also
appear in the GROPU BY clause.
SELECT COUNT(*) AS cnt, name, comment, ...
FROM tst
GROUP BY name, comment, ...
HAVING COUNT(*) = 1;
Is the requirement of
johnf [EMAIL PROTECTED] writes:
On Sunday 27 January 2008 10:56:18 am Mike Ginsburg wrote:
Each of columns that you specify in your SELECT clause, must also
appear in the GROPU BY clause.
Is the requirement of select fields matching group by fields a SQL92
requirement or something to due to
Mike,
I can't do that with my comments - I get all six of the records in the
result with the example instead of just four like I want . . but someone
else had a solution without using the group by clause . .
Phil.
On Sun, 2008-01-27 at 13:56 -0500, Mike Ginsburg wrote:
Hi Phil,
Each of
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