i have a table like this
ID nachname vorname
1 john doe
2 john doe
3 jim doe
4 Michael Knight
I need a query that will return all the fields (select *) from the records that have the same nachname and vorname (in this case, records 1 and 2). Can anyone help me with this? Thanks
The following query will give the list of duplicates :
SELECT n1.* FROM table n1
inner join table n2 on n2.vorname=n1.vorname and n2.nachname=n1.nachname
where n1.id <> n2.id
BTW The data you posted seems to be wrong "Doe" and "Knight" are a lastname, not a firstname :p.
The general solution to your problem is a query of the form
SELECT col1, col2, count(*)
FROM t1
GROUP BY col1, col2
HAVING count(*) > 1
This will return one row for each set of duplicate row in the table. The last column in this result is the number of duplicates for the particular values.
If you really want the ID, try something like this:
SELECT id FROM
t1,
( SELECT col1, col2, count(*)
FROM t1
GROUP BY col1, col2
HAVING count(*) > 1 ) as t2
WHERE t1.col1 = t2.col1 AND t1.col2 = t2.col2
Haven't tested it though
You can do it with a self-join:
select distinct t1.id from t as t1 inner join t as t2
on t1.col1=t2.col1 and t1.col2=t2.col2 and t1.id<>t2.id
the t1.id<>t2.id
is necessary to avoid ids matching against themselves. (If you want only 1 row out of each set of duplicates, you can use t1.id<t2.id
).
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