SELECT * FROM products p LEFT JOIN criteria_values cv ON p.key=cv.key LEFT JOIN
criteria c ON cv.key=c.key WHERE c.value IS NOT NULL
Hard to answer without more detail, but I am guessing the answer will be
something like the above. Your question makes it hard to understand whether c
or cv is
SELECT * FROM products p LEFT JOIN criteria_values cv ON p.key=cv.key LEFT
JOIN criteria c ON cv.key=c.key WHERE c.value IS NOT NULL
Hard to answer without more detail, but I am guessing the answer will be
something like the above. Your question makes it hard to understand whether
c or cv is
I'm going to jump in and throw in my 2 cents...
Have you used dreamweaver?
I would suggest Dreamweaver to any new programmer beginning php/mysql.
It helped me out tremendously in the beginning. I'm not an advanced programmer
with hand coding classes yet, but I can get any job completed for
Since we are just tossing out development environments.
We moved to Aptana in conjunction with TortoiseSVN for a team environment
development timelines dropped.
Personally I do not feel any gui editor makes you a better programmer, maybe
you understand the fundamentals a little less.
Not that
On Sat, 20 Nov 2010 13:54:29 -0700
Ben Miller biprel...@gmail.com wrote:
Hi,
I'm building a website for a client in which I need to compare their
products, side-by-side, but only include criteria for which all selected
products have a value for that criteria.
In my database (MySQL), I
I'm trying to get the total number of a certain records from a database,
but the result is always '1'. Please advise!
=MySql Table =
=activitiy =
id | employee_id | project_id | date
1 | 45 | 60 | 2003-09-09
2 | 34 | 10 | 2003-09-10
3 | 45 | 45
Chris Kay wrote:
The query does not error out it just does not give any records, and I
Know
What part of The query does not error out do you not understand.
Why are there so many people willing to say what is wrong with a code but when it
comes to
A solution that go silent.
I find
This is a MySQL question and best directed to the MySQL mailing lists
available at:
http://www.mysql.com/documentation/lists.html
-Original Message-
From: Chris Kay [mailto:[EMAIL PROTECTED]]
Sent: Thursday, June 13, 2002 4:33 PM
To: PHP General List
Subject: [PHP] MySQL Query
ps... from a PHP perspective, you may find troubleshooting things like this
easier by using formatting like this:
$sql =
select
detail.*,
type.type_name,
status.status_name,
staff.staff_name,
Man, where do I start. There could be so many things wrong. First of
all, this is a PHP list, not MySQL. Second, use MySQL_error() after you
issue a query to see if an error was returned
http://www.php.net/mysql_error. Third, in this line:
, 14 June 2002 1:26 PM
To: Chris Kay; 'PHP General List'
Subject: RE: [PHP] MySQL Query Help
Man, where do I start. There could be so many things wrong.
First of all, this is a PHP list, not MySQL. Second, use
MySQL_error() after you issue a query to see if an error was
returned
On Fri, 14 Jun 2002, Chris Kay wrote:
The query does not error out it just does not give any records, and
I
Know
What part of The query does not error out do you not understand.
Why are there so many people willing to say what is wrong with a
code
but when it comes to A solution
[snip]
select cust_fnn, cust_name, agroup_access.group_access_cust from cust,
agroup_access where
agroup_access.group_access_group='$id'
cust.cust_fnn!=agroup_access.group_access_cust order by cust.cust_name
[/snip]
try this (note syntactical differences);
select cust_fnn, cust_name,
On Mon, Sep 10, 2001 at 03:59:36PM -0500, Sheridan Saint-Michel wrote:
Well, I played with this a little more and it seems to be acting oddly when
you first
call this select unless you set the variable first. So if the below doesn't
work try
actually doing this
$query=set @count=NULL;
See if this does what you are shooting for:
select tableName.*,if(@count=1,@count:=@count+1,@count:=1) as inc, from
tableName;
Keep in mind that @count will keep it's value until the thread is closed, so
if for some reason you have
to do the above twice in one thread throw in a statement like:
www.foxjet.com
- Original Message -
From: Sheridan Saint-Michel [EMAIL PROTECTED]
To: Michael George [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Monday, September 10, 2001 3:29 PM
Subject: Re: [PHP] MySQL query help
See if this does what you are shooting for:
select tableName.*,if(@count=1
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