RE: [PHP] MySQL problem
| From: | Simon Kimber | Date: | Wed, 04 Jul 2001 23:28:24 +0000 |
| Subject: | RE: [PHP] MySQL problem | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-56238@lists.php.net to get a copy of this message | ||
Sorry!!! I'm stupid! I forgot to mention that the list of causes has to be
for a specified accident_report.weekending
Cheers
Simon
> -----Original Message-----
> From: Don Read [mailto:dread@texas.net]
> Sent: 04 July 2001 23:21
> To: Simon Kimber
> Cc: php-general@lists.php.net
> Subject: RE: [PHP] MySQL problem
>
>
>
> On 04-Jul-01 Simon Kimber wrote:
> > Hi All,
> >
> > Does anyone know if this can be done with one query?
> >
> > I have to create a chart based on info in two tables that are
> four tables
> > apart.
> >
> > Here are the relevant tables and just the most relevant fields...
> >
> > accident_report
> > - ID
> > - weekending (this is a YYYY-MM-DD format date)
> > - (and others)
> >
> > accident_data
> > - ID
> > - accident_report_id
> > - (and others)
> >
> > accident_cause (a lookup table)
> > - ID
> > - accident_data_id
> > - cause_id
> >
> > cause (a list of possible causes of accidents ie. "falling object" or
> > "electric shock"
> > - ID
> > - Description
> >
> >
> > I need to list all the causes with the number of times each has
> occurred,
> > even if it's zero times... they don't need to be listed in any
> particular
> > order...
> >
>
> "select cause.ID, count(*) as cnt from ...
> WHERE ...
> group by cause.ID";
>
> Regards,
> --
> Don Read dread@texas.net
> -- It's always darkest before the dawn. So if you are going to
> steal the neighbor's newspaper, that's the time to do it.
>