问题
So I have a table Car(Car_id, Value, Make)
, another table Person(Person_id, Name)
and a third table linking the two together PerCar(Car_id, Person_id)
.
So I have this correlated subquery:
SELECT MAKE, VALUE
FROM CAR WHERE VALUE > (SELECT AVG(VALUE) FROM CAR C WHERE C.CAR_ID = CAR_ID);
So if I wanted to display the person's name also, is this also valid correlated sub query:
SELECT Car.Make, Car.Value, Person.Name
FROM PerCar NATURAL JOIN Car
NATURAL JOIN Person where Car.Value > (SELECT AVG(VALUE) FROM Car);
回答1:
Your query should be as below-
You can check DEMO HERE
SELECT Car.Make,
Car.Value,
Person.Name
from Car
INNER JOIN CarPer
ON Car.Car_id = CarPer.Car_id
INNER JOIN Person
ON Person.Person_id = CarPer.Person_ID
WHERE Car.Value > (SELECT AVG(VALUE) FROM Car);
Remember, here joining can be INNER, OUTER, LEFT or other joins based on your requirement and data structure.
回答2:
This is NOT a correlated subquery:
SELECT MAKE, VALUE
FROM CAR
WHERE VALUE > (SELECT AVG(VALUE) FROM CAR C WHERE C.CAR_ID = CAR_ID);
Always qualify all column references in a query.
You think the WHERE
clause is:
WHERE CAR.VALUE > (SELECT AVG(VALUE) FROM CAR C WHERE C.CAR_ID = CAR.CAR_ID);
But it is really:
WHERE CAR.VALUE > (SELECT AVG(C.VALUE) FROM CAR C WHERE C.CAR_ID = C.CAR_ID);
There is no reference to the outer query, so this is not correlated. I would write this as:
SELECT C.MAKE, C.VALUE
FROM CAR C
WHERE C.VALUE > (SELECT AVG(C2.VALUE) FROM CAR C2 WHERE C2.CAR_ID = C2.CAR_ID);
However, it is unlikely that this will return any rows. Given your lack of problem statement and the values in your tables, a sensible query would return "all cars whose value is higher than the average for their make"
SELECT C.MAKE, C.VALUE
FROM CAR C
WHERE C.VALUE > (SELECT AVG(C2.VALUE) FROM CAR C2 WHERE C2.MAKE = C.MAKE);
Because the rest of your question is predicated on the first query being a "correlated subquery", the rest of the question does not make sense.
来源:https://stackoverflow.com/questions/59124475/is-this-a-valid-correlated-sub-query-with-joins