Re: [PHP-DB] HELP With UPDATE Query in mySQL

2003-10-06 Thread pete M 
and put it inside a transaction cos in case something goes wrong part of 
the way through

pete

Jeff Shapiro wrote:

This should do it:

UPDATE Bookings
SET
Booking_Start_Date = CONCAT(DATE_FORMAT(Booking_Start_Date, 
'%Y-%m-%d'), '09:00:00'),
Booking_End_Date = CONCAT(DATE_FORMAT(Booking_End_Date, '%Y-%m-%d'), 
'17:30:00');

If you are using version 4.1.1 or newer the DATE_FORMAT function could 
be replaced by:
DATE(Booking_Start_Date)

On Fri, 3 Oct 2003 22:54:49 +0100, Shaun spoke thusly about [PHP-DB] 
HELP With UPDATE Query in mySQL:

Hi,

I have two columns in my Bookings table of type DATETIME -
Booking_Start_Date and Boking_End_Date. How can i update every row so that
all of the times for Booking_Start_Date are 09.00 and all of the times for
Booking_End_Date are 17.30, without affecting any of the dates?
Thanks for your help


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Re: [PHP-DB] HELP With UPDATE Query in mySQL

2003-10-04 Thread Jeff Shapiro 

This should do it:

UPDATE Bookings
SET
Booking_Start_Date = CONCAT(DATE_FORMAT(Booking_Start_Date, 
'%Y-%m-%d'), '09:00:00'),
Booking_End_Date = CONCAT(DATE_FORMAT(Booking_End_Date, '%Y-%m-%d'), 
'17:30:00');

If you are using version 4.1.1 or newer the DATE_FORMAT function could 
be replaced by:
DATE(Booking_Start_Date)


On Fri, 3 Oct 2003 22:54:49 +0100, Shaun spoke thusly about [PHP-DB] 
HELP With UPDATE Query in mySQL:
> Hi,
> 
> I have two columns in my Bookings table of type DATETIME -
> Booking_Start_Date and Boking_End_Date. How can i update every row so that
> all of the times for Booking_Start_Date are 09.00 and all of the times for
> Booking_End_Date are 17.30, without affecting any of the dates?
> 
> Thanks for your help

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Jeff Shapiro

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Re: [PHP-DB] Help with update...

2001-08-18 Thread BD 

it's amazing.. I've been banging my head for over an hour on this, then as
soon as I send for help, I see the obvious - first of all, my error checking
was done *before* the query executed... then, to make it even more of an
effort in mindlessness, the table name capitalization was throwing it all
off!!

Sorry for the waste of good bandwidth...

~BD~



http://www.bustdustr.net
http://www.rfbdproductions.com
Home Of Radio Free BD
For The Difference.


- Original Message -
From: BD <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Saturday, August 18, 2001 12:21 PM
Subject: [PHP-DB] Help with update...


> I'm working under the gun here, so please excuse me if this turns out to
be
> something really simple, but it's got me stumped... I'm trying to update
the
> "cover" field in a table by using the album_id field in the table and
adding
> a ".jpg" extention to it...
> For example:
> "Update albums set cover = "1.jpg" where album_id = 1"
>
> For some reason, however, the update isn't working when executed... here
are
> the results I get:
>
> album_id: 1 - cover: 1.jpg
> UPDATE ALBUMS SET COVER = "1.jpg" WHERE ALBUM_ID = 1
> mysql_errno: 0:   mysql_error:
> Update Failed!
>
> As you can see, there's no error code generated, and the update statement
> works great from the command line...
>
> here's the relevant code (i'm connecting to the database as root, so i
don't
> think it's a permissions thing...):
> $query = "SELECT * FROM albums ORDER BY album_id";
>
> $stuff = mysql_query($query) or die("Select Failed!");
>
> while ($results = mysql_fetch_array($stuff)) {
>
>   $cover = $results['album_id'].".jpg";
>   $album_id = $results['album_id'];
>   echo $album_id." - ".$cover."";
>   $update = "UPDATE ALBUMS SET COVER = \"$cover\" WHERE ALBUM_ID =
> $album_id";
>   echo $update."";
>   echo mysql_errno().": ".mysql_error()."";
>   mysql_query($update) or die("Update Failed!");
>  }
>
>
> Simple... right? But it's beating me up this morning.. I would greatly
> appreciate it if anyone could show me the error of my ways...
>
> ~BD~
>
> http://www.bustdustr.net
> http://www.rfbdproductions.com
> Home Of Radio Free BD
> For The Difference.
>
>
>
>
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