Question

I am having some trouble with a query and was hoping someone could help. I have tried searching for a solution, but can’t seem to find a similar scenario with my search terms.

Here is what I am looking for:

I have a table with three columns and need to return the values in the first column, and two counts of the values in the second column. One count is based on a parameter and the other is the total count. Ideally, I would only like to return the values in column one that have equal counts.

For example:

Part Number | Make ID
ABC123      | 1<br>
ABC123      | 1<br>
ABC123      | 3<br>
DEF456      | 1<br>
DEF456      | 1

Part Number | Count of Apps Where Make ID = 1| Count of Total Apps
ABC123      |                              2 | 3
DEF456      |                              2 | 2

The query I have so far will return the total count of apps for part numbers that have the parameter value, but I need it to return both counts:

SELECT apps.part#, 
       COUNT(DISTINCT application#)apps 
FROM   [mytable] AS apps 
       INNER JOIN (SELECT part# 
                   FROM   [mytable] 
                   WHERE  make = '1') AS sz 
               ON sz.part# = apps.part# 
GROUP  BY apps.part# 
ORDER  BY 1 

Any help is greatly appreciated!

Thanks, everyone! I received several correct answers, and selected the one that was give first:

SELECT part#, 
       COUNT(DISTINCT CASE WHEN make = '1' THEN application# END) make1_apps,
       COUNT(DISTINCT application#) total_apps
FROM   [mytable]
GROUP  BY part#
HAVING COUNT(DISTINCT CASE WHEN make = '1' THEN application# END)
     = COUNT(DISTINCT application#)
ORDER  BY part#

Thanks again!

Answer 1

SELECT part#, 
       COUNT(DISTINCT CASE WHEN make = '1' THEN application# END) make1_apps,
       COUNT(DISTINCT application#) total_apps
FROM   [mytable]
GROUP  BY part#
HAVING COUNT(DISTINCT CASE WHEN make = '1' THEN application# END)
     = COUNT(DISTINCT application#)
ORDER  BY part#; 

The HAVING clause ensures the two counts match.