On Tue, Feb 15, 2011 at 12:19 PM, Peter Lind wrote:
> On 15 February 2011 20:28, Ethan Rosenberg wrote:
>> Dear List -
>>
>
> Ask google you should
> Plenty of advice you'll find
> we won't do your homework
>
> Regards
> Peter
>
+1. Ethan, you should read this [1] first.
Regards,
Tommy
[1] ht
On 15 February 2011 20:28, Ethan Rosenberg wrote:
> Dear List -
>
> I have a form. In one field, the customer types the name of a product.
> The first seven(7) results of the MySQL query that the entry generates
> should be displayed as a clickable drop down list.
>
> How do I do it?
Ask googl
Ah, I see. In Brad's reply there was two $result = mssql_query($sql) or
die(mssql_error()); in the code. Removed the one from outside of the loop
and it works fine now.
Thanks to both of you for your help!
On 1/18/07, Chris <[EMAIL PROTECTED]> wrote:
Dan Shirah wrote:
> The code above d
Dan Shirah wrote:
The code above displays no information at all.
What I want to do is:
1. Retrieve my information
2. Assign it to a variable
3. Output the data into a table with each unique record in a seperate row
As Brad posted:
echo "";
while ($row = mssql_fetch_array($result)) {
$id
The code above displays no information at all.
What I want to do is:
1. Retrieve my information
2. Assign it to a variable
3. Output the data into a table with each unique record in a seperate row
On 1/18/07, Brad Bonkoski <[EMAIL PROTECTED]> wrote:
Dan Shirah wrote:
> Hello all,
>
> I am tr
Dan Shirah wrote:
Hello all,
I am trying to pull data and then loop through the multiple results
display
in seperate rows.
My database contains several tables which are all tied together by the
credit_card_id. After running the query, it ties the unique record
together
by matching the credi
ent: Tuesday, April 16, 2002 9:58 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP] Displaying Results
On Wednesday 17 April 2002 08:21, Jason Soza wrote:
> Sorry, I just noticed that the count() function will do at least the
> first part of my question, i.e. SELECT year, COUNT(*) FROM cars GROUP
Sent: Wednesday, April 17, 2002 10:21 AM
To: [EMAIL PROTECTED]
Subject: Re: [PHP] Displaying Results
Sorry, I just noticed that the count() function will do at least the
first part of my question, i.e. SELECT year, COUNT(*) FROM cars GROUP
BY year
But the second part still has me a bit stump
On Wednesday 17 April 2002 08:21, Jason Soza wrote:
> Sorry, I just noticed that the count() function will do at least the
> first part of my question, i.e. SELECT year, COUNT(*) FROM cars GROUP
> BY year
>
> But the second part still has me a bit stumped. I know that you can
> pass a variable usi
You're part way there:
SELECT COUNT(matches) FROM table GROUP BY matches;
- Original Message -
From: Jason Soza <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Tuesday, April 16, 2002 5:49 PM
Subject: [PHP] Displaying Results
Suppose I want a script that goes into a table, looks up a
Sorry, I just noticed that the count() function will do at least the
first part of my question, i.e. SELECT year, COUNT(*) FROM cars GROUP
BY year
But the second part still has me a bit stumped. I know that you can
pass a variable using something like script.php?year=1991, but doesn't
that as
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