I am running this query on MySQL:
CREATE TABLE DuplicateSKU
SELECT * FROM FeedsAll
INNER JOIN (
SELECT FeedsAll.SKU
FROM FeedsAll
GROUP BY FeedsAll.SKU
HAVING COUNT(*) > 1) as DuplicateSKU;
and it is giving this error:
#1248 - Every derived table must have its own alias
What is wrong?
Help, please!
You are using the name DuplicateSKU twice, once for the new table once for your subquery. These should be different names, e.g.
create table duplicatesku
select *
from feedsall
inner join
(
select sku
from feedsall
group by sku
having count(*) > 1
) as duplicates on duplicates.sku = feedsall.sku;
By the way you were cross-joining (missing ON clause, which MySQL doesn't report unfortunately). I added the appropriate ON clause.
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 have a complex query which results in a table which includes a time column. There are always two rows with the same time:
The result also contains a value column. The value of two rows with the same time is always different.
I now want to extend the query to join the rows with the same time together. So my thought was to join the derived table like this:
SELECT A.time, A.value AS valueA, B.value as valueB FROM
(
OLD_QUERY
) AS A INNER JOIN A AS B ON
A.time=B.time AND
A.value <> B.value;
However, the JOIN A AS B part of the query does not work. A is not recognized as the derived table. MySQL is searching for a table A in the database and does not find it.
So the question is: How can I join a derived table?
You cannot join a single reference to a table (or subquery) to itself; a subquery must be repeated.
Example: You cannot even do
SELECT A.* FROM sometable AS A INNER JOIN A ...
The A after the INNER JOIN is invalid unless you actually have a real table called A.
You can insert the subquery's results into another table, and use that; but it cannot be a true TEMPORARY table, as those cannot be joined to themselves or referenced twice at all in almost any query. _By referenced twice, I mean joined, unioned, used as an "WHERE IN" subquery when it is already referenced in the FROM.
If nothing else distinguishes the rows, you can just use aggregation to get the two values:
select time, min(value), max(value)
from (<your query here>) a
group by time;
In MySQL 8+, you can use a cte:
with a as (
<your query here>
)
select a1.time, a1.value, a2.value
from a a1 join
a a2
on a1.time = a2.time and a1.value <> a2.value;
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).
Please see the image below:
Given the original table, I need to create the derived table in MySQL on the server.
With CREATE TABLE ... AS SELECT DISTINCT... I am able to create the derived table with the Category column, but trying in vain to create the Category_count column. Can you kindly point out how to solve this?
Not very comfortable with MySQL or SQL even, hence the request.
Many thanks in advance!
Use the following if the table does not exists:
CREATE TABLE derived AS
SELECT Category, COUNT(*) AS Category_count
FROM original
GROUP BY Category
Use the following if the table already exists and without creating duplicate rows:
INSERT INTO derived (Category, Category_count)
SELECT Category, COUNT(*) AS Category_count
FROM original o
WHERE NOT EXISTS (SELECT * FROM derived d WHERE d.Category = o.Category)
GROUP BY Category
Use the following to update the rows already exists:
UPDATE derived LEFT JOIN (
SELECT Category, COUNT(*) AS Category_count
FROM original
GROUP BY Category
)x ON derived.Category = x.Category
SET derived.Category_count = x.Category_count
Select catoregy,count(id)as category_count into derived from original group by catoregy
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).