Thanks, but how can I pass the current users value from the other query ?
On Thu, Feb 3, 2011 at 4:22 PM, Simcha Younger wrote:
> On Thu, 3 Feb 2011 13:55:36 +
> Tompkins Neil wrote:
>
> > SELECT DISTINCT(away_teams_id) AS teams_id
> > FROM fixtures_results
> > WHERE (fixtures_results.away_
On Thu, 3 Feb 2011 13:55:36 +
Tompkins Neil wrote:
> SELECT DISTINCT(away_teams_id) AS teams_id
> FROM fixtures_results
> WHERE (fixtures_results.away_users_id = *users.users_id*)
> Any ideas why I'm getting Unknown column 'users.users_id' in 'where clause'
> for the part of the statement th
Hi Frederico,
the precedence between the comma-operator and JOIN changed
with 5.0.12.
See http://dev.mysql.com/doc/refman/5.0/en/join.html
Excerpt from that article:
Previously, the comma operator (,) and JOIN both had the same
precedence, so the join expression t1, t2 JOIN t3 was interpreted as
Try not mixing left join and comma-joins, and use an INNER JOIN keyword
between "m.manufacturers_id, products_to_categories"
Baron
Federico Giannici wrote:
Since we upgraded from MySQL 4.0 to 5.0 (under OpenBSD 4.1 amd64) the
following command:
select count(*) as total from products_descript
Sorry:-\ Meant to say
I do not understand why mySQL is pointing to this as an error
Bob
-Original Message-
From: Bartis, Robert M (Bob) [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 01, 2006 5:30 PM
To: 'mysql'
Subject: Unknown column 'testcase_root.Test' in 'order clause'
I am
akul,
Wednesday, May 01, 2002, 6:30:06 AM, you wrote:
a> SELECT countries.*, lnk0.Value AS fraudIndex, lnk1.Value AS enabled
a> FROM o as countries LEFT JOIN l_TINYINT AS lnk0 ON
a> lnk0.FID=countries.ID AND lnk0.TID=55 LEFT JOIN l_TINYINT AS lnk1 ON
a> lnk1.FID=countries.ID AND lnk1.TID=57 WHERE
yan Shrout
>
> -Original Message-
> From: Bill "Elvis" Gibbs [mailto:[EMAIL PROTECTED]]
> Sent: Monday, June 18, 2001 3:20 PM
> To: Shrout, Ryan
> Subject: RE: Unknown column in 'where clause' error
>
>
> WHERE Customer_No = 'CMET355853923
gem original-
De: Shrout, Ryan [mailto:[EMAIL PROTECTED]]
Enviada: Segunda-feira, 18 de Junho de 2001 20:14
Para: '[EMAIL PROTECTED]'
Assunto: RE: Unknown column in 'where clause' error
Thanks for the suggestions, but this is actually a variable that is passed
via a URL. For e
***
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> -Original Message-
> From: Shrout, Ryan [mailto:[EMAIL PROTECTED]]
> Sent: 18 June 2001 20:14
> To: '[EMAIL PROTECTED]'
> Subjec
ustomer_No = '".$id."';
if for some reason the first one didn't work.
HTH,
Cal
*
* Cal Evans
* Senior Internet Dreamer
* http://www.calevans.com
*
- Original Message -
From: "Shrout, Ryan" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday
Hello,
Your field(column) is a CHAR and therefore the parameter to it should be in
single
quotes:
Change your query to:
SELECT * FROM customers WHERE Customer_No = 'CMET3558539239'
*
* Visit http://www.computerstaff.net - Compu
uotes around' a variable?
Ryan Shrout
-Original Message-
From: Bill "Elvis" Gibbs [mailto:[EMAIL PROTECTED]]
Sent: Monday, June 18, 2001 3:20 PM
To: Shrout, Ryan
Subject: RE: Unknown column in 'where clause' error
WHERE Customer_No = 'CMET3558539239'
you nee
Try
SELECT * FROM customers WHERE Customer_No = 'CMET3558539239'
Jorge Oliveira
[EMAIL PROTECTED]
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-Mensagem original-
De: Shrout, Ryan [mail
Put quotes around the Customer_No:
SELECT * FROM customers WHERE Customer_No = 'CMET3558539239'
- TIM
> Here is the full error I am getting when running the query included:
>
> Failed Query: SELECT * FROM customers WHERE Customer_No = CMET3558539239
> 1054 : Unknown column 'CMET3558539239' in
On Mon, Jun 18, 2001 at 02:44:00PM -0400, Shrout, Ryan wrote:
> Here is the full error I am getting when running the query included:
>
> Failed Query: SELECT * FROM customers WHERE Customer_No = CMET3558539239
> 1054 : Unknown column 'CMET3558539239' in 'where clause'
> Here's the problem, I know
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