Here's an example (snip) from a var_dump of that
$BuildPerUniqueDateArray:
(note that the 'aweber_7solar_aw' table does NOT have a record for
the
date '2009-07-28', so I would expect to see that "1" to be a "0"
there.)
I fixed this. The problem was the way I was building the array inside
Here's an example (snip) from a var_dump of that
$BuildPerUniqueDateArray:
(note that the 'aweber_7solar_aw' table does NOT have a record for
the
date '2009-07-28', so I would expect to see that "1" to be a "0"
there.)
If a table doesn't have a record for a given date, I wouldn't expect
t
> Here's an example (snip) from a var_dump of that $BuildPerUniqueDateArray:
> (note that the 'aweber_7solar_aw' table does NOT have a record for the
> date '2009-07-28', so I would expect to see that "1" to be a "0" there.)
If a table doesn't have a record for a given date, I wouldn't expect to s
take a look at this:
http://otn.oracle.com/oramag/oracle/04-mar/o24asktom.html
and search for the "Analytics to the Rescue" example. Instead of 3 seconds
you want 1800 and instead of sum you want count.Don't forget to group by
ip,of course... And you're done. No need for an extra table.
Hop
Whatabout creating a table containing online users,
where you log every activity with IP, BrowserSession and Timestamp.
You also create a table to track the accual unique visits.
So my logic to solve it :
Update the online table like this (Some rough coding below, not tested at
all, read the logi
In article <[EMAIL PROTECTED]>,
[EMAIL PROTECTED] says...
> What is the Aliases used for???
>
> > Aliases. SELECT COUNT(*) AS howmany FROM table
> >
> > Then use the variable $howmany
At this stage I refer you to The Fine (mysql) Manual - or anything on the
SQL language.
--
David Robley
Temp
Mike
- Original Message -
From: "Ron Allen" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, April 11, 2002 7:13 AM
Subject: [PHP-DB] Re: count
> I know how to get the results for the total number of records
> select count(*) from $table
&g
What is the Aliases used for???
> Aliases. SELECT COUNT(*) AS howmany FROM table
>
> Then use the variable $howmany
>
> --
> David Robley
> Temporary Kiwi!
>
> Quod subigo farinam
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
In article <[EMAIL PROTECTED]>,
[EMAIL PROTECTED] says...
> I know how to get the results for the total number of records
> select count(*) from $table
> but how do I put them into a variable for me to use later???
>
> "David Robley" <[EMAIL PROTECTED]> wrote in message
> [EMAIL PROTECTED]">news
I know how to get the results for the total number of records
select count(*) from $table
but how do I put them into a variable for me to use later???
"David Robley" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> In article <[EMAIL PROTECTED]>,
> [EMAIL PROTEC
In article <[EMAIL PROTECTED]>,
[EMAIL PROTECTED] says...
> I would like to count all the total records from a database. How would I go
> about doing that???
Do a COUNT * on each table and total the results? Although the results
may actually not mean much depending on your DB structure. Or do
Hi Barry!
you can do it like this for example:
$query = "SELECT * FROM artist WHERE artist_name LIKE 'b%' ORDER BY artist
ASC";
$count = mysql_query("SELECT COUNT(artist) AS count FROM artist WHERE
artist_name LIKE 'b%'",$db);
$x = mysql_fetch_array($count);
$result = mysql_query($query) or
Thank you everyone. The "COUNT(*) AS c" worked great.
---Original Message---
From: Kai Voigt
Date: Friday, 25
January 2002 11:11:06 a.
To: Barry Rumsey
Cc: [EMAIL PROTECTED]
Subject: Re:
Count(*)
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