I've given my derived table an alias but still when i run the query it gives me the error "Every derived table must have its own alias".
select a,b,c,sum(d) as 'sum'
from(select a,b,c,sum(d)as 'd' from e join f using(z)) as 'alias'
group by a;
EDIT: better sample
----This gives derived table error-----
select name,sum(pop) as 'total' from(select name as 'name',sum(population)
as pop from table1 join table2 using(countrycode));
----This gives me the SQL syntax error--------
select name,sum(pop) as 'total' from(select name as 'name',sum(population)
as pop from table1 join table2 using(countrycode)) as 'alias';
Thanks in advance.
Try this :
select name,
sum(pop) as `total`
from(
select name as `name`, sum(population) as pop
from table1
join table2 using(countrycode)
) `m` <---- alias needed here too
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 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).