Re: difficult count statement

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What you can do:

- create an auxiliary table months( char(2) not null )
- feed it with values '01', ..., '12'
- left join table 'months' with your result table, grouped by month. Then
you will have all months.

It is always handy to have a few such auxiliary tables in your DB, like
"years, "months", "weeks", "year_weeks", "ordinals", ...

HTH
Ignatius
____________________________________________
----- Original Message -----
From: "Merlin" <news.groups@web.de>
To: <php-db@lists.php.net>
Sent: Monday, May 05, 2003 5:32 PM
Subject: Re:  difficult count statement


> > SELECT     DATE_FORMAT( date_column, '%Y- %m- %d') AS mydate,
> >                    COUNT(*) AS c
> > FROM         table
> > GROUP BY mydate
>
> This wokes. Thank you ignatius! I have worked on the charts and it helped
me
> to get a lot further.
> There is a point where I still have not found a good solution.
>
> Problem is, that there are some days where no data has changed (for
example
> a day with no new registered members).
> In this case the query should return '0'
>
> E.g.:
> Jan    10
> Feb   32
> Mar   0
> Apr   5
>
> At the moment I would not have march listed in my chart which causes
> problems.
>
> E.g.:
> Jan    10
> Feb   32
> Apr   5
>
> Do u have an idea on how to include the 0?
>
> Thanx for any help on that.
>
> merlin
>
>
>
> --
> Worldwide Travel Community
> http://www.globosapiens.net
> "Ignatius Reilly" <ignatius.reilly@free.fr> schrieb im Newsbeitrag
> 091a01c311a6$73f70e00$0100a8c0@server">news:091a01c311a6$73f70e00$0100a8c0@server...
> > Then you may try:
> >
> > SELECT     DATE_FORMAT( date_column, '%Y- %m- %d') AS mydate,
> >                    COUNT(*) AS c
> > FROM         table
> > GROUP BY mydate
> >
> > Ignatius
> > ____________________________________________
> > ----- Original Message -----
> > From: "Merlin" <news.groups@web.de>
> > To: <php-db@lists.php.net>
> > Sent: Saturday, May 03, 2003 8:23 PM
> > Subject: Re:  difficult count statement
> >
> >
> > > > SELECT date_column, COUNT(*) AS c FROM table GROUP BY date_column
> > > >
> > > > Will give you a count for each date in the table.
> > >
> > > Thank you. This works great. There is just one important prblem. This
> > > coulumn contains datetime. So it cant be grouped to get the amounts of
> reg
> > > users for each day. It gives me always 1 for a specific time.
> > >
> > > E.g:
> > > date:2002-07-29 17:21:00members:1
> > > date:2002-07-29 17:21:50members:1
> > > date:2002-07-29 17:23:05members:1
> > >
> > > I played around now for 1/2 hour and could not figure out how to
change
> > this
> > > into what I want. Can you give me some more help on this?
> > >
> > > Thank you in advance,
> > >
> > > Merlin
> > >
> > >
> > >
> > > --
> > > Worldwide Travel Community
> > > http://www.globosapiens.net
> > > "John W. Holmes" <holmes072000@charter.net> schrieb im Newsbeitrag
> > > 000801c3101f$59d88ae0$7c02a8c0@coconut">news:000801c3101f$59d88ae0$7c02a8c0@coconut...
> > > > > I am running a mysql db and I would like to figure out how to
count
> > > > member
> > > > > growth.
> > > > >
> > > > > For a chart I would need the number of members who have registered
> > > > each
> > > > > day.
> > > > >
> > > > > Something like this:
> > > > > count (*) AS c
> > > > > WHERE
> > > > >  regdate = '20030501'
> > > > >
> > > > > does not really make sense, since I would have to query the db for
> > > > each
> > > > > day
> > > > > making it dramaticly slow.
> > > > > Has anybody an idea on how to get the numbers registered each day
to
> > > > fead
> > > > > a
> > > > > graph with data?
> > > >
> > > > SELECT date_column, COUNT(*) AS c FROM table GROUP BY date_column
> > > >
> > > > Will give you a count for each date in the table. If you want it for
a
> > > > range, then use:
> > > >
> > > > SELECT date_column, COUNT(*) AS c FROM table WHERE date_column
BETWEEN
> > > > 20030101 AND 20030201 GROUP BY date_column
> > > >
> > > > Substitute your own dates, of course.
> > > >
> > > > ---John W. Holmes...
> > > >
> > > > PHP Architect - A monthly magazine for PHP Professionals. Get your
> copy
> > > > today. http://www.phparch.com/
> > > >
> > > >
> > >
> > >
> > >
> > > --
> > > PHP Database Mailing List (http://www.php.net/)
> > > To unsubscribe, visit: http://www.php.net/unsub.php
> > >
> > >
> >
>
>
>
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>


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