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
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 have a tables named products,criteria and
criteria_values
If I have something like
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
-Original Message-
From: Ben Miller [mailto:biprel...@gmail.com]
Sent: Saturday, November 20, 2010 3:54 PM
To: 'php-general'
Subject: [PHP] MySQL Query Help
Hi,
I'm building a website for a client in which I need to compare their
products, side-by-side, but only include criteria
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
Hello everyone!
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
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
I have a rather longer query which I would like to get all records past todays date.
Here is my query
$ttwo = date(YmdGi);
$dbq = select(select detail.*, type.type_name, status.status_name, staff.staff_name,
source.source_long,
source.source_short from detail, type, status, staff,
source
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
;
$dbq = select($sql);
-Original Message-
From: Chris Kay [mailto:[EMAIL PROTECTED]]
Sent: Thursday, June 13, 2002 4:33 PM
To: PHP General List
Subject: [PHP] MySQL Query Help
I have a rather longer query which I would like to get all
records past todays date.
Here
: Chris Kay [mailto:[EMAIL PROTECTED]]
Sent: Thursday, June 13, 2002 7:33 PM
To: PHP General List
Subject: [PHP] MySQL Query Help
I have a rather longer query which I would like to get all records
past
todays date.
Here is my query
$ttwo = date(YmdGi);
$dbq = select(select detail
, 14 June 2002 1:26 PM
To: Chris Kay; 'PHP General List'
Subject: RE: [PHP] MySQL Query Help
Man, where do I start. There could be so many things wrong.
First of all, this is a PHP list, not MySQL. Second, use
MySQL_error() after you issue a query to see if an error was
returned
On Fri, 14 Jun 2002, Chris Kay wrote:
The query does not error out it just does not give any records, and
I
Know
What part of The query does not error out do you not understand.
Why are there so many people willing to say what is wrong with a
code
but when it comes to A solution
I have a query
select cust_fnn, cust_name, agroup_access.group_access_cust from cust,
agroup_access where
agroup_access.group_access_group='$id'
cust.cust_fnn!=agroup_access.group_access_cust order by cust.cust_name
The 2 tables are as follows
agroup_access
agroup_access_id
[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;
I'm trying to make a query that will number it's own output rows. e.g. when
listing all the entries in a table that are related to a specific invoice,
there will be a column with a monotonically increasing integer value (1-x
where x is the number of matching entries).
I know I can easily do
:
set @count=Null;
between the selects =P
Sheridan Saint-Michel
Website Administrator
FoxJet, an ITW Company
www.foxjet.com
- Original Message -
From: Michael George [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Monday, September 10, 2001 2:20 PM
Subject: [PHP] MySQL query help
I'm
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
21 matches
Mail list logo
