Solved

# finding records in table using the count()

Posted on 2014-07-28
114 Views
I have an attendance table in sql where I store data for activities by organizations.  Each record in my attendancegrid is associated to a specific month and fiscal year.

I am looking to get all the Distinct records based on ID from my table that have a total > 0 throughout the whole year.  I would like to group it by activity and committee.

Here is what I am using.  Does it look like I am on the right track using the count() function?

``````SELECT committee, Activity, Count(Distinct ID) as NumberofParticipants FROM AttendanceGrid where Total > 0
and Month IN ('July', 'August', 'September', 'October', 'November','December','January', 'February','March','April', 'May','June')
And Fiscal = 2014 GROUP BY Committee, Activity order by Committee, Activity
``````
0
Question by:al4629740
[X]
###### Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

• Help others & share knowledge
• Earn cash & points
• 4
• 2
• 2
• +1

LVL 8

Assisted Solution

5teveo earned 250 total points
ID: 40224828
move 'Distinct'

SELECT Distinct  committee, Activity, Count(ID) as NumberofParticipants FROM AttendanceGrid where Total > 0
and Month IN ('July', 'August', 'September', 'October', 'November','December','January', 'February','March','April', 'May','June')
And Fiscal = 2014 GROUP BY Committee, Activity order by Committee, Activity
0

LVL 1

Expert Comment

ID: 40225006
Rest of the query seems ok, you can remove month condition (as you are looking for all 12 months data)
0

Author Comment

ID: 40225296
Why would I move distinct? Doesn't that make committee distinct if I did that?
0

Author Comment

ID: 40225578
What I want is distinct IDs of the students that had any totals.  I tested the change and believe that my original still is the right one.

0

LVL 8

Expert Comment

ID: 40225796
Do you want to show the detail ID# numbers and to show the counts of grouped committee and activities? Both at the same time? If yes - then these are reporting issues and not data selection issues.

What does a result record look like? provide an example.
0

Author Comment

ID: 40239143
Here is what worked

``````SELECT r.Sector, count(Distinct r.RegID)
FROM tblOrgRegistrations AS r
LEFT JOIN tblOrgHours AS h
ON h.RegID = r.RegID
AND Month(ActivityDate) = 7 And r.Board = 1
LEFT JOIN tblOrgActivities AS a
ON h.ActivityID = a.ActivityID And h.Hours > 0
LEFT JOIN tblOrgProfile AS o
ON h.AgencyID = o.AgencyID Where o.AgencyID = 74 And a.ActivityName = 'Board Meeting'
Group By r.Sector order by r.Sector
``````
0

LVL 75

Accepted Solution

Anthony Perkins earned 250 total points
ID: 40239987
That looks familiar and I am sorry to see that you still incorrectly using a LEFT JOIN.  In other words, this:
LEFT JOIN tblOrgActivities AS a ON h.ActivityID = a.ActivityID  AND h.Hours > 0
and this:
WHERE   ...
AND a.ActivityName = 'Board Meeting'
Means that you are in fact doing an (implicit) INNER JOIN and not a LEFT JOIN.

But you can confirm this by inspecting the results and you will see that all the a.ActivityName are "Board Meeting".  If that is what you really want than you should change the LEFT JOIN to an INNER JOIN

Also, aliasing all your columns is strongly suggested (see ActivityDate), so that you are not the only one that can read the query,
0

Author Comment

ID: 40254442
Can you explain further how you determined I was doing an implicit INNER JOIN from this section?

LEFT JOIN tblOrgActivities AS a ON h.ActivityID = a.ActivityID  AND h.Hours > 0
and this:
WHERE   ...
AND a.ActivityName = 'Board Meeting'
Means that you are in fact doing an (implicit) INNER JOIN and not a LEFT JOIN.
0

LVL 75

Expert Comment

ID: 40254706
Can you explain further how you determined I was doing an implicit INNER JOIN from this section?
The best way to explain this is for you to test it out for yourself.  Here is a very simple example:
``````DECLARE @Table1 TABLE (
ID integer NOT NULL PRIMARY KEY,
Item varchar(20))

DECLARE @Table2 TABLE (
ID integer NOT NULL PRIMARY KEY,
ItemType varchar(20))

INSERT	@Table1 (ID, Item)
VALUES	(1, 'Item 1'),
(2, 'Item 2'),
(3, 'Item 3'),
(4, 'Item 4'),
(5, 'Item 5'),
(6, 'Item 6'),
(7, 'Item 7'),
(8, 'Item 8')

INSERT	@Table2 (ID, ItemType)
VALUES	(2, 'Item Type 1'),
(3, 'Item Type 1'),
(5, 'Item Type 2'),
(8, 'Item Type 2')

-- 1
SELECT *
FROM	@Table1 t1
INNER JOIN @Table2 t2 ON t1.ID = t2.ID
WHERE	t2.ItemType = 'Item Type 2'

-- 2
SELECT *
FROM	@Table1 t1
LEFT JOIN @Table2 t2 ON t1.ID = t2.ID
WHERE	t2.ItemType = 'Item Type 2'

-- 3
SELECT *
FROM	@Table1 t1
LEFT JOIN @Table2 t2 ON t1.ID = t2.ID AND t2.ItemType = 'Item Type 2'
``````

Note how the results from Query 1 and 2 are the same, even though 2 is written as an LEFT  JOIN,  If you really want a LEFT JOIN than the correct way to write it, is as Query 3.

If you are still not convinced, look at the Execution plan (press Ctrl+M before executing F5) for all three queries.  Again Query 1 and 2 are identical.  Note how they both read as Nested Loops (Inner Join) and the last one as Nested Loops (Left Outer Join) .
0

## Featured Post

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

### Suggested Solutions

Ever needed a SQL 2008 Database replicated/mirrored/log shipped on another server but you can't take the downtime inflicted by initial snapshot or disconnect while T-logs are restored or mirror applied? You can use SQL Server Initialize from Backupâ€¦
In the first part of this tutorial we will cover the prerequisites for installing SQL Server vNext on Linux.
Familiarize people with the process of utilizing SQL Server functions from within Microsoft Access. Microsoft Access is a very powerful client/server development tool. One of the SQL Server objects that you can interact with from within Microsoft Acâ€¦
Using examples as well as descriptions, and references to Books Online, show the different Recovery Models available in SQL Server and explain, as well as show how full, differential and transaction log backups are performed
###### Suggested Courses
Course of the Month4 days, 16 hours left to enroll