From memory earlier versions of MySQL don't support subselects. The
latest (and greatest) does so you may want to check the version you are
using - I can't remember when it was introduced, I have since switched
to PostGreSQL...
all the best,
graeme.
Juffermans, Jos wrote:
Hi,
I'm afraid I can't
Hi,
I'm afraid I can't help you with the query - I'm used to Oracle which as a
bigger set of SQL commands.
I can help you with your PHP question. If you want to put a variable in a
query, you can add the variable directly into the string:
$sql_query = "select zg_longitude from zip_code wh
Hello,
on 04/05/2005 11:58 AM Hendy Irawan said the following:
Would anyone be kind enough to provide me some pointers about object
relationship mapping [libraries] in PHP?
All I know is there are some DB APIs available: PEAR DB, MDB, ADOdb,
Creole... I currently use MDB for its excellent DB schema
All,
I have a question as how I can return some results that I am looking
for. Ideally I'd like to run this sql statement but for some reason MySql
won't allow me to run it:
SELECT
*
FROM
zip_code
WHERE
zg_latitude between ((select zg_latitude from zip_code where zg_zipcode =
'78730')
http://www.weberdev.com/get_example-4063.html to load the db
http://www.weberdev.com/get_example-4062.html to get them out
bastien
From: "Noel Simela" <[EMAIL PROTECTED]>
To:
Subject: [PHP-DB] Loading images on MYSQL through a PHP web page
Date: Wed, 6 Apr 2005 22:50:48 +0200
Hie Guys
Does anyone
Hie Guys
Does anyone know how to load pictures through php to a mysql db...i want
users on my site to be able upload their pictures on the db,whenever i
view their profiles i should be able to see their pics
regards
Noel
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Hi There (again),
My apologies for being a complete pest with this issue. I changed a
few things around and I can now pass the image name thanks to Jos,
however, I'm getting a permission errors now.
Warning: imagejpeg(): Unable to open 'corneilus.jpg' for writing
I've changed image folder and s
Hi,
Are you trying to display more images at once then?
You can store the image like you do now, then have the GD script output the
image tag and use the image you've just stored as a source.
GD script:
...
ImageJPEG($thumb, $route_photo, '100');
echo '';// add this line
ImageDestroy($thumb)
Hi Jos,
I understand what your saying. The problem is all the image names are
listed in the DB (image.jpg, image1.jpg and so on) and all images have
titles attached. For example,
Title: Large Green Sea Turtle
Image: LargeGreenSeaTurtle.jpg
As you can see, the GD script randomly pulls photo n
Hi,
If I understand correctly, you're using
"include/function_gd.php?route_photo=$thumb" as image source which is the
script you pasted. If you use a script as image source, the script should
output the image data - not just store the image.
I'm not sure what you're trying to do but if you have a
I believe the problem is somewhere in the GD script. Because when I
view the source it shows an empty variable. I should at least see the
name of variable in the source. Any ideas?
On Apr 6, 2005, at 9:46 AM, Bastien Koert wrote:
Have a look here...I have some sample code on calling images
Have a look here...I have some sample code on calling images from a db. It
used the two pages like you do here and passes the ID back and forth to the
image production page...maybe it will help
http://www.weberdev.com/get_example-4062.html
bastien
From: Craig Hoffman <[EMAIL PROTECTED]>
To: php-
Hi,
You could do something like this:
SELECT 'Accounts' as department, Accounts as id FROM mytable WHERE Accounts
IS NOT NULL
UNION ALL
SELECT 'Human Resources' as department, HR as id FROM mytable WHERE HR IS
NOT NULL
UNION ALL
SELECT 'Support' as department, Support as id FROM my
that's a fine idea!
I'll hve a play...
Cheers...
Martin Norland <[EMAIL PROTECTED]>
06/04/2005 14:51
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php-db@lists.php.net
To
php-db@lists.php.net
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Subject
Re: [PHP-DB] find a value in entire table...
[EMAIL PROTECTED] wrote:
> I need to search an entire table for
[EMAIL PROTECTED] wrote:
I need to search an entire table for a value, and then report back what
field it was found in... how on earth do I do that?
I've a list of departments, as field names.
whenever a user interacts with that Dpet, I wanna add thier id No to the
appropriate field.
[snip]
So I
Thanks everyone for answering my other post. However, I'm still having
trouble passing the image name from the GD script to the image tag.
I'm including the GD script below. Any help would be great. Thanks -
Craig
//GD script
include ("db.php");
$query = "SELECT route_photo FROM routes WHE
I need to search an entire table for a value, and then report back what
field it was found in... how on earth do I do that?
I've a list of departments, as field names.
whenever a user interacts with that Dpet, I wanna add thier id No to the
appropriate field.
so I'll be left with a table that loo
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