This sample also work in PhpMyAdmin, but not from the php code.
 Please advise.

 **see previous post, for sample code**

 Thanks Rick.

 -john

=P e p i e  D e s i g n s
 www.pepiedesigns.com
 Providing Solutions That Increase Productivity

 Web Developement. Database. Hosting. Multimedia.

On Tue, 21 Jan 2003, Rick Emery wrote:

> SELECT risk_level, COUNT(*) FROM mytable GROUP BY risk_level;
>
> ----- Original Message -----
> From: "[-^-!-%-" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Tuesday, January 21, 2003 6:06 AM
> Subject: [PHP] ADV SQL Help Needed.
>
>
>
> Hello everyone,
>
> I need help with the following query.
>
> I have a table with a column named 'RISK_LEVEL'.
> RISK_LEVEL can hold three values: Low, Moderate, and High.
>
> I need a query that counts the number of occurences of each option.
> Now, I know I can do
>
> 'select count(risk_level) as levelCount where risk_level=low'
>
> for each level (low, moderate, and high). But, I was wondering if there's
> anyway to merge the queries into one.
>
> I'm looking for something like
>
> "select count(risk_level='low') as lowCount, count(risk_level='moderate')as
> modCount, count(risk_level='high') as highCount "
>
> Which will return, lowCount = 1 , modCount=3, and highCount=1, for table
> with
>
>  id | risk_level
>  1  | low
>  2  | moderate
>  3  | moderate
>  4  | high
>  5  | moderate
>
> How do you do it? Can tha be done?
>
> When I run the code above, I get the same number for each record. i.e.
> lowCount, modCount, and highCount all come out to 5--which is the number
> of records in the table.
>
> Is this possible, or should I just run multiple queries?
> Please help.
>
>
> =P e p i e  D e s i g n s
>  www.pepiedesigns.com
>  Providing Solutions That Increase Productivity
>
>  Web Developement. Database. Hosting. Multimedia.
>
>
>
>
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