Doh! Just saw the "..orderid equals the tagid in both the.." Make that:
select tagnumber,days,expdate from tags
left join membership on membership.tagid = tags.tagid
where orderid = membership.tagid
there are a few other ways to do it. The most efficient depends on which
table has the most records, how many orderids match the tagid in the same
table .. etc...
On Wed, Sep 05, at 07:26pm, Ian Moore ([EMAIL PROTECTED]) wrote:
> There may be a better way, but right off the top of my head I'd say:
>
> select tagnumber,days,expdate from membership
> left join tags on membership.tagid = tags.tagid
> where tags.tagid is not null
>
>
> On Wed, Sep 05, at 06:10pm, [EMAIL PROTECTED] ([EMAIL PROTECTED]) wrote:
>
> > Hello All,
> >
> > I got a sql query I seemed to be missing the obvious or just can't see.
> > Database Table structure:
> > CREATE TABLE `membership` (
> > `tagid` int(8) NOT NULL default '0',
> > `startdate` date NOT NULL default '0000-00-00',
> > `expdate` date NOT NULL default '0000-00-00',
> > `days` smallint(4) NOT NULL default '0',
> > PRIMARY KEY (`tagid`),
> > KEY `days` (`days`)
> > )
> >
> > CREATE TABLE `tags` (
> > `tagid` int(8) NOT NULL auto_increment,
> > `orderid` int(8) NOT NULL default '0',
> > `finderid` int(8) NOT NULL default '0',
> > `tagnumber` varchar(10) NOT NULL default '',
> > PRIMARY KEY (`tagid`),
> > UNIQUE KEY `tagnumber` (`tagnumber`),
> > KEY `orderid` (`orderid`,`finderid`)
> > );
> >
> > What I need to do is extract all the tagnumbers,days and expdate where the
> > orderid equals the tagid in both the "tags" and "membership" tables.
> >
> > I just can't seem to get the results I need. Any help would be much appreciated.
> >
> > Mysql version 3.23.40
> >
> > THX's
> > Mike(mickalo)Blezien
> > =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
> > Thunder Rain Internet Publishing
> > Providing Internet Solutions that work!
> > http://www.thunder-rain.com
> > Tel: 1(225)686-2002
> > =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
> >
> >
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