On Sun, 21 Nov 2010 11:19:04 -0700
"Ben Miller" wrote:
>
> To help clarify - the 3 tables look something like the following (tableName
> => column,column,column...):
>
> Products => product_id,product_name,product_description... (key =
> product_id)
> Criteria => criteria_id,criteria_title,cri
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 joi
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 jo
On Sat, 20 Nov 2010 13:54:29 -0700
"Ben Miller" 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 have a tables
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
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 client
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.
> 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 come
PROTECTED]]
> Sent: Friday, 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_e
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:
'detail_start_date_y,detail_start_date_m,detail_start_dat
ps... from a PHP perspective, you may find troubleshooting things like this
easier by using formatting like this:
$sql = "
select
detail.*,
type.type_name,
status.status_name,
staff.staff_name,
source.source_
This is a MySQL question and best directed to the MySQL mailing lists
available at:
http://www.mysql.com/documentation/lists.html
> -Original Message-
> From: Chris Kay [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, June 13, 2002 4:33 PM
> To: PHP General List
> Subject: [PHP] MySQL Query
[snip]
"select cust_fnn, cust_name, agroup_access.group_access_cust from cust,
agroup_access where
agroup_access.group_access_group='$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, agroup
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=N
xJet, an ITW Company
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 th
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:
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