Tek-Tips is the largest IT community on the Internet today!

Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!

  • Congratulations TouchToneTommy on being selected by the Tek-Tips community for having the most helpful posts in the forums last week. Way to Go!

2 Tables... The Dogz World

Status
Not open for further replies.

slurpyx

Programmer
Aug 1, 2002
59
PH
This is the scenario:

cmp_table
id description thecalldate
1 Pit Bull 2004-04-25 02:03:20
2 Stray Dog 2004-04-25 20:21:32

etc..
etc....

$query = "select year(thecalldate) as 'year',DAYOFMONTH(thecalldate) as 'monthday',DAYNAME(thecalldate) as 'weekname', MONTHNAME(thecalldate) as 'monthname', count(*) as 'no' from cmp_table where month(thecalldate) = '$draw_m' and year(thecalldate) = '$draw_y' group by dayofmonth(thecalldate) order by thecalldate";

while($myrow=mysql_fetch_array($lookresult)){
$no=$myrow["no"];
$callyear=$myrow["year"];
$callmonth=$myrow["monthname"];
$dayname=$myrow["weekname"];
$monthday=$myrow["monthday"];

echo "$callmonth $monthday, $callyear - $dayname ------ $no";
}

this in return would produce...

April 25, 2004 - Thursday ------ 2


Now my boss tells me that we need to add all the dogs.. i mean disabled dogs! and keep it in diff column when reporting...


my boss provided me with this table (with data)...

inc_table
id description datetime
1 1 Eyed Pit Bull 2004-04-25 02:03:20
2 3 Legged Stray Dog 2004-04-25 20:21:32




i have 2 tables (cmp_table and inc_table)
my question now is... how can i create a qry that would result an output like this:

DATE complete dog incomplete dog
April 25, 2004 - Thursday 2 4



I need advice on this.. or my head is going to land on a slinger!
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor

Back
Top