I want to do something like this:
SELECT *
FROM db.table
WHERE COUNT(someField) > 1
How can I achieve this in MySql?
Use the HAVING
, not WHERE
clause, for aggregate result comparison.
Taking the query at face value:
SELECT *
FROM db.table
HAVING COUNT(someField) > 1
Ideally, there should be a GROUP BY
defined for proper valuation in the HAVING
clause, but MySQL does allow hidden columns from the GROUP BY...
Is this in preparation for a unique constraint on someField
? Looks like it should be...
It should also be mentioned that the "pk" should be a key field. The self-join
SELECT t1.* FROM db.table t1
JOIN db.table t2 ON t1.someField = t2.someField AND t1.pk != t2.pk
by Bill Karwin give you all the records that are duplicates which is what I wanted. Because some have more than two, you can get the same record more than once. I wrote all to another table with the same fields to get rid of the same records by key fields suppression. I tried
SELECT * FROM db.table HAVING COUNT(someField) > 1
above first. The data returned from it give only one of the duplicates, less than 1/2 of what this gives you but the count is good if that is all you want.
As OMG Ponies stated, the having clause is what you are after. However, if you were hoping that you would get discrete rows instead of a summary (the "having" creates a summary) - it cannot be done in a single statement. You must use two statements in that case.
I give an example up on Group By between two table in Sql:
Select cn.name,ct.name,count(ct.id) totalcity
from city ct left join country cn on ct.countryid = cn.id
Group By cn.name,ct.name
Having totalcity > 2
One way
SELECT t1.*
FROM db.table t1
WHERE exists
(SELECT *
FROM db.table t2
where t1.pk != t2.pk
and t1.someField = t2.someField)
Here you go:
SELECT Field1, COUNT(Field1)
FROM Table1
GROUP BY Field1
HAVING COUNT(Field1) > 1
ORDER BY Field1 desc