# subquery – What is the error “Every derived table must have its own alias” in MySQL?

## The Question :

407 people think this question is useful

I am running this query on MySQL

SELECT ID FROM (
SELECT ID, msisdn
FROM (
SELECT * FROM TT2
)
);



and it is giving this error:

Every derived table must have its own alias.

What’s causing this error?

• Couldn’t you just simplify this as “select ID from TT2”?
• I got this error recently because I had an extra ) in a query with a lot of UNION ALLs.
• Seeing as how this is the #1 Google search… The accepted answer doesn’t really answer the error ‘Every derived table must have its own alias’. Look below for more info.

572 people think this answer is useful

Every derived table (AKA sub-query) must indeed have an alias. I.e. each query in brackets must be given an alias (AS whatever), which can the be used to refer to it in the rest of the outer query.

SELECT ID FROM (
SELECT ID, msisdn FROM (
SELECT * FROM TT2
) AS T
) AS T



In your case, of course, the entire query could be replaced with:

SELECT ID FROM TT2



78 people think this answer is useful

I think it’s asking you to do this:

SELECT ID
FROM (SELECT ID,
msisdn
FROM (SELECT * FROM TT2) as myalias
) as anotheralias;



But why would you write this query in the first place?

18 people think this answer is useful

Here’s a different example that can’t be rewritten without aliases ( can’t GROUP BY DISTINCT).

Imagine a table called purchases that records purchases made by customers at stores, i.e. it’s a many to many table and the software needs to know which customers have made purchases at more than one store:

SELECT DISTINCT customer_id, SUM(1)
FROM ( SELECT DISTINCT customer_id, store_id FROM purchases)
GROUP BY customer_id HAVING 1 < SUM(1);



..will break with the error Every derived table must have its own alias. To fix:

SELECT DISTINCT customer_id, SUM(1)
FROM ( SELECT DISTINCT customer_id, store_id FROM purchases) AS custom
GROUP BY customer_id HAVING 1 < SUM(1);



( Note the AS custom alias).

1 people think this answer is useful

I arrived here because I thought I should check in SO if there are adequate answers, after a syntax error that gave me this error, or if I could possibly post an answer myself.

OK, the answers here explain what this error is, so not much more to say, but nevertheless I will give my 2 cents using my words:

This error is caused by the fact that you basically generate a new table with your subquery for the FROM command.

That’s what a derived table is, and as such, it needs to have an alias (actually a name reference to it).

So given the following hypothetical query:

SELECT id, key1
FROM (
SELECT t1.ID id, t2.key1 key1, t2.key2 key2, t2.key3 key3
FROM table1 t1
LEFT JOIN table2 t2 ON t1.id = t2.id
WHERE t2.key3 = 'some-value'
) AS tt



So, at the end, the whole subquery inside the FROM command will produce the table that is aliased as tt and it will have the following columns id, key1, key2, key3.

So, then with the initial SELECT from that table we finally select the id and key1 from the tt.