Counting a row in sql

Hey,

I have to display a column that is a count of another row on another table. I have done it using two queries but I would like to find out how to do it in one query.

The column to display will be "Number of Exercises" and it will count the number of "Exercise Id's" on the Session Details table.

I have attached my code and the tables to help you all understand how they are related.

Thanks.
$sql = "Select Sessions.Date_Created, Sessions.Created_By
        From Sessions, Session_Details
        Where Sessions.Member_Id = '$clientID'";
    $result = mysql_query($sql);
    while ($ROW = mysql_fetch_array($result)) {
        $sessionID = $ROW['Session_Id'];
		$sessionDate = $ROW['Date_Created'];
        echo date('F d, Y',strtotime($sessionDate));
        echo $createdBy = $ROW['Created_By'];
	}
//    Retrieve the number of exercises in that session
    $sql2 = "Select Session_Details.Exercise_Id
        From Session_Details, Sessions
        Where Sessions.Member_Id = '$clientID' AND Session_Details.Session_Id = Sessions.Session_Id";
    $result2 = mysql_query($sql2);
    echo mysql_num_rows($result2);

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tables.bmp
tqrecordsAsked:
Who is Participating?
 
Guy Hengel [angelIII / a3]Connect With a Mentor Billing EngineerCommented:
simply add a join
Select s.Date_Created, s.Created_By, count(sd.exercice_id) exercice_count
 , m.username created_by_user
From Sessions s 
left join Session_Details sd
  on sd.session_id = s.session_id
join members m
  on m.member_id = s.created_by
Where s.Member_Id = '$clientID'
group by s.Member_id

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Guy Hengel [angelIII / a3]Billing EngineerCommented:
you mean:
Select s.Date_Created, s.Created_By, count(sd.exercice_id) exercice_count
From Sessions s 
left join Session_Details sd
  on sd.session_id = s.session_id
Where s.Member_Id = '$clientID'
group by s.Member_id

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0
 
tqrecordsAuthor Commented:
Thanks, but there is also one more join I need to make from another table.

There is a created_by column in the Sessions table and it is an integer value.

The created_by will match the member_id on the members table and will return the name of the member.

I have the updated tables attached.


tables.bmp
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