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).
Related
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 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 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 have the following table:
CREATE TABLE sometable (my_id INTEGER PRIMARY KEY AUTOINCREMENT, name STRING, number STRING);
Running this query:
SELECT * FROM sometable;
Produces the following output:
1|someone|111
2|someone|222
3|monster|333
Along with these three fields I would also like to include a count representing the amount of times the same name exists in the table.
I've obviously tried:
SELECT my_id, name, count(name) FROM sometable GROUP BY name;
though that will not give me an individual result row for every record.
Ideally I would have the following output:
1|someone|111|2
2|someone|222|2
3|monster|333|1
Where the 4th column represents the amount of time this number exists.
Thanks for any help.
You can do this with a correlated subquery in the select clause:
Select st.*,
(SELECT count(*) from sometable st2 where st.name = st2.name) as NameCount
from sometable st;
You can also write this as a join to an aggregated subquery:
select st.*, stn.NameCount
from sometable st join
(select name, count(*) as NameCount
from sometable
group by name
) stn
on st.name = stn.name;
EDIT:
As for performance, the best way to find out is to try both and time them. The correlated subquery will work best when there is an index on sometable(name). Although aggregation is reputed to be slow in MySQL, sometimes this type of query gets surprisingly good results. The best answer is to test.
Select *, (SELECT count(my_id) from sometable) as total from sometable
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).