Grouping counts from tables by account ID
I took the liberty of throwing this in to an SQLFiddle here

If I play with the query, and run the screened and approved queries on the data that I chose, I see you have a condition which may or may not be a bug. If you consider my data, where I have multiple accountId values per application, then, your subquery:


SELECT      application.accountId, 0, COUNT(*), 0
FROM        application
JOIN        termsofapproval ON application.accountId = termsofapproval.accountId
JOIN        approvaluserjoin ON termsofapproval.id = approvaluserjoin.termsofapprovalId
[accountIdCondition]    
GROUP BY    application.accountId



That query will duplicate the count of approved users if there are multiple applications for the same accountId. In the SQLFiddle I have, it returns a count of 4 for only 2 distinct approvaluserjoin records.

It is likely that in your data it is not possible to get that condition though... right?

Regardless. I believe the more logical representation of your query is as follows (which I have in this SQLFiddle here

SELECT application.accountId ,
       count(distinct application.id) as ApplicantsCount,
       count(distinct approvaluserjoin.id) as ApprovedCount,
       count(distinct screened.id) as ScreenedCount
FROM application
left join termsofapproval on application.accountId = termsofapproval.accountId
left join approvaluserjoin on approvaluserjoin.termsofapprovalId = termsofapproval.id
left join screened on screened.applicationId = application.id
where application.accountId &lt;= 2
group by application.accountId


Note how there is only one join, using left outer joins. Also note that a count of a null value is 0, so the null values in the outer-join results do not contribute to the sum. The coutn(distinct ...) construct allows you to count the things you are interested in, even if the query returns them in multiple contexts.

You will need to carefully understand the query, the implications are different to yours, and it may be more accurate than what you have (or less accurate).