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How do I query a table and get the ranking of records based off of multiple ranked records?

Posted on 2011-03-13
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Last Modified: 2012-05-11
I have a table that has several records of ranks. The ranking can be 0-5.

So my table is like so...

record_ID |  project_ID  |  ranking  |  date |

1. record ID is just Auto Incremented
2. project_ID is a random number and can have multiple records of the same Project_ID
3. ranking can be 0-5 (0 = no points, 5 = Highest points).
4. And the date

I need for the query to COMBINE the records with the same project_ID and ORDER each combined record by the AVERAGE ranking (5 being first).

I've written some basic/intermediate SQL queries but this one is giving me some trouble.

Thank you,

Brian
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Question by:brihol44
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7 Comments
 
LVL 5

Expert Comment

by:mayankagarwal
ID: 35125526
select avg(ranking), project_id from mytable group by project_id
0
 
LVL 39

Expert Comment

by:Pratima Pharande
ID: 35125529
did you mean that you want one record for each project with average ranking ??

like
Project Id  ranking

then

Select Project_Id , Sum(ranking)/count(ranking) from Tablename
group by project_id

if not
give sampe data and show us what result you need
0
 
LVL 5

Expert Comment

by:mayankagarwal
ID: 35125535
sorry forgot the order by clause

select avg(ranking) as avg_rank, project_id from mytable group by project_id order by avg_rank desc
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Author Comment

by:brihol44
ID: 35125676
Sorry but writing it out actually made it more clear for directions. I'll remember to do this in the future sorry if it's way off from what I was originally asking.

Brian
sc.jpg
0
 
LVL 2

Accepted Solution

by:
sihar86 earned 500 total points
ID: 35125781
SELECT project_ID, SUM(ranking), COUNT(1)*5
FROM tablename
WHERE ranking > 0
GROUP BY project_ID

Open in new window


since 0 would not count,
I add where ranking > 0
0
 
LVL 39

Expert Comment

by:Pratima Pharande
ID: 35125783
try this
Select Project_Id , Sum(ranking) as with_point ,(count(Project_Id) *5 ) as out_of_point  from Tablename
group by project_id

Correction in your result
for project 3 have 2 records

so 5 out of 10
0
 
LVL 5

Expert Comment

by:mayankagarwal
ID: 35125887
SELECT project_ID, SUM(ranking), COUNT(project_ID)*5, date
FROM tablename
WHERE ranking > 0
GROUP BY project_ID
0

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