Johan Höök schrieb:
Hi Barry,
see: http://dev.mysql.com/doc/refman/5.0/en/insert-select.html
you cannot insert into a table you're doing select on
(same goes for update).
But i am doing it on a test server version 5.x and it works like a charm :)
--
Smileys rule (cX.x)C --o(^_^o)
Dance for me!
Hi Barry,
see: http://dev.mysql.com/doc/refman/5.0/en/insert-select.html
you cannot insert into a table you're doing select on
(same goes for update).
Regards,
/Johan
Barry skrev:
I get this error:
Fehler in
/home/virtual/site4/fst/var/www/html/adminheaven/artikel-vererben-save.php
in
I get this error:
Fehler in
/home/virtual/site4/fst/var/www/html/adminheaven/artikel-vererben-save.php
in Zeile 36 mit Error:
Query:INSERT INTO objektflyer_verknuepfung
(av_o_id_haupt,av_o_id_link,av_text,av_op_id) SELECT
418,av_o_id_link,av_text,av_op_id FROM objektflyer_verknuepfung WHERE
a
Please replace YES_COLUMN with the column name that stores the 'Yes' values.
Best regards,
Diego Wald
- Original Message -
From: "Shaun" <[EMAIL PROTECTED]>
To:
Sent: Wednesday, November 23, 2005 11:21 AM
Subject: INSERT SELECT Problem
Hi,
The following qu
ct_ID =".$project_id)
Met vriendelijke groet,
Almar van Pel
-Oorspronkelijk bericht-
Van: Shaun [mailto:[EMAIL PROTECTED]
Verzonden: woensdag 23 november 2005 15:22
Aan: mysql@lists.mysql.com
Onderwerp: INSERT SELECT Problem
Hi,
The following query worked fine:
INSERT IN
Hi,
The following query worked fine:
INSERT INTO Allocations(Project_ID, User_ID)
SELECT P.Project_ID, U.User_ID
FROM Users U, Projects P, Clients C
WHERE P.Client_ID = C.Client_ID
AND U.Client_ID = C.Client_ID
AND Project_ID =".$project_id)
However I want to add a column to the INSERT part of
Hello ,
I'm working on a project with MySql 4.0.12-log.
I have a problem with insert .. select:
To describe the program of some touristic tours I create two tables:
TOUR that contains the data
TAPPE that contains the program of the tour.
(relation 1:n).
To keep track of each tour i create two tab
I am performing a query in two tables and output it into a temporary table
for further sorting.
The problem is that in one of the tables (TableB), I have an additional id
that I want to be added to the temptable. This id is not present in TableA.
I get this error message when I run the query for T