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?
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?
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
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?
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).
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 own 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).
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
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
.
Then, with the initial SELECT
, we finally select the id
and key1
from that generated table (tt
).
I noticed that there were some threads with similar questions, and I did look through them but did not really get a convincing answer. Here s my question: The subquery below returns a Table with 3 ...
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