I got an error: #1242 — Subquery returns more than 1 row when i run this sql.
CREATE VIEW test
AS
SELECT cc_name,
COUNT() AS total,
(SELECT COUNT(*)
FROM bed
WHERE respatient_id > 0
GROUP BY cc_name) AS occupied_beds,
(SELECT COUNT(*)
FROM bed
WHERE respatient_id IS NULL
GROUP BY cc_name) AS free_beds
FROM bed
GROUP BY cc_name;
OMG Ponies
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asked Oct 23, 2011 at 16:00
6
The problem is that your subselects are returning more than one value — IE:
SELECT ...
(SELECT COUNT(*)
FROM bed
WHERE respatient_id IS NULL
GROUP BY cc_name) AS free_beds,
...
…will return a row for each cc_name, but SQL doesn’t support compacting the resultset for the subselect — hence the error.
Don’t need the subselects, this can be done using a single pass over the table using:
SELECT b.cc_name,
COUNT(*) AS total,
SUM(CASE
WHEN b.respatient_id > 0 THEN 1
ELSE 0
END) AS occupied_beds,
SUM(CASE
WHEN b.respatient_id IS NULL THEN 1
ELSE 0
END) AS free_beds
FROM bed b
GROUP BY b.cc_name
answered Oct 23, 2011 at 16:05
OMG PoniesOMG Ponies
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8
This is because your subqueries (the SELECT bits that are inside parentheses) are returning multiple rows for each outer row. The problem is with the GROUP BY; if you want to use subqueries for this, then you need to correlate them to the outer query, by specifying that they refer to the same cc_name as the outer query:
CREATE VIEW test
AS
SELECT cc_name,
COUNT() AS total,
(SELECT COUNT()
FROM bed
WHERE cc_name = bed_outer.cc_name
AND respatient_id > 0) AS occupied_beds,
(SELECT COUNT(*)
FROM bed
WHERE cc_name = bed_outer.cc_name
WHERE respatient_id IS NULL) AS free_beds
FROM bed AS bed_outer
GROUP BY cc_name;
(See http://en.wikipedia.org/wiki/Correlated_subquery for information about correlated subqueries.)
But, as OMG Ponies and a1ex07 say, you don’t actually need to use subqueries for this if you don’t want to.
answered Oct 23, 2011 at 16:11
ruakhruakh
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2
Your subqueries return more than 1 row. I think you you need something like :
SELECT COUNT(*) AS total,
COUNT(CASE WHEN respatient_id > 0 THEN 1 END) AS occupied_beds,
COUNT(CASE WHEN respatient_id IS NULL THEN 1 END) AS free_beds
FROM bed
GROUP BY cc_name
You can also try to use WITH ROLLUP + pivoting (mostly for learning purposes, it’s a much longer query ) :
SELECT cc_name,
MAX(CASE
WHEN num_1 = 1 THEN tot_num END) AS free_beds,
MAX(CASE
WHEN num_1 = 2 THEN tot_num END) AS occupied_beds,
MAX(CASE
WHEN num_1 = IS NULL THEN tot_num END) AS total
FROM
(SELECT cc_name, CASE
WHEN respatient_id > 0 THEN 1
WHEN respatient_id IS NULL THEN 2
ELSE 3 END as num_1,
COUNT(*) as tot_num
FROM bed
WHERE
CASE
WHEN respatient_id > 0 THEN 1
WHEN respatient_id IS NULL THEN 2
ELSE 3 END != 3
GROUP BY cc_name,
num_1 WITH ROLLUP)A
GROUP BY cc_name
answered Oct 23, 2011 at 16:06
a1ex07a1ex07
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SELECT COUNT()
FROM bed
WHERE respatient_id > 0
GROUP BY cc_name
You need to remove the group-by in the sub query, so possibly something like
SELECT COUNT(*)
FROM bed
WHERE respatient_id > 0
or possibly — depending on what your application logic is….
SELECT COUNT(*) from (
select count(*),cc_name FROM bed
WHERE respatient_id > 0
GROUP BY cc_name) filterview
answered Oct 23, 2011 at 16:02
SorenSoren
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Здравствуйте, не понимаю какое добавить еще условие, чтобы решить
DELIMITER //
CREATE TRIGGER Buget_Trigger
AFTER INSERT
ON Buget5
FOR EACH ROW BEGIN
if((select MaxCount from Buget5 where (select max(dateee) where dateee<(now())) )<(select Buget from Buget5 where (select max(dateee) where dateee<(now())))) then
update Insects set Count=Count+20 where(select max(datee) where datee<(now()));
update Mammals set Count=Count+20 where(select max(datee) where datee<(now()));
update ColdBloodedness set Count=Count+20 where(select max(datee) where datee<(now()));
end if;
END //
insert into Mammals(mammals_id, id, typee, count, datee) VALUES (1,null,'',7,'2019-12-13');
insert into ColdBloodedness(ColdBloodedness_Id, id, typee,datee, count) VALUES (1,null,'','2019-12-13',10);
insert into Insects(insects_id, id, typee, count,datee) VALUES (1,null,'',5, '2019-12-13');
insert into Buget5(buget_id, id, buget, maxcount, dateee) VALUES (1, null,600, 200,'2019-12-13');
select * from Insects;
select * from ColdBloodedness;
select * from Mammals;Но если сделаю второй раз, то ничего не сработает и будет ошибка 1242
insert into Buget5(buget_id, id, buget, maxcount, dateee) VALUES (2, null,600, 200,'2019-12-13');
select * from Insects;
select * from ColdBloodedness;
select * from Mammals;I am trying to make a select statement that selects the image names from a MySQL database.
The table is called — pictures_archive. I am also trying to select these pictures depending on the category they have. The code is:
SELECT pictures_archive_filename FROM pictures_archive
WHERE pictures_archive_id = (SELECT pictures_archive_id
FROM pictures_archive_category WHERE pictures_category_id = 9)
It gives me an «#1242 — Subquery returns more than 1 row» error. I can see why, but can’t figure it out how to do it.
Marc Alff
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asked Sep 26, 2012 at 8:30
Since your subquery can return multiple values, IN should fit in youy where clause.
SELECT pictures_archive_filename
FROM pictures_archive
WHERE pictures_archive_id IN
(
SELECT pictures_archive_id
FROM pictures_archive_category
WHERE pictures_category_id = 9
)
an alternative for this is to join both tables which is more efficient.
SELECT pictures_archive_filename
FROM pictures_archive a
INNER JOIN pictures_archive_category b
ON a.pictures_archive_id = b.pictures_archive_id
WHERE b.pictures_category_id = 9
answered Sep 26, 2012 at 8:32
John WooJohn Woo
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2
Use IN instead of equal (=)
SELECT pictures_archive_filename FROM pictures_archive
WHERE pictures_archive_id IN (SELECT pictures_archive_id
FROM pictures_archive_category WHERE pictures_category_id = 9)
OR if possible use a JOIN between 2 tables
answered Sep 26, 2012 at 8:31
PrasannaPrasanna
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SELECT pictures_archive_filename FROM pictures_archive
WHERE pictures_archive_id = (SELECT pictures_archive_id
FROM pictures_archive_category WHERE pictures_category_id = 9 LIMIT 1)
answered Sep 26, 2012 at 8:33
PermGenErrorPermGenError
46k8 gold badges87 silver badges106 bronze badges
SELECT p.pictures_archive_filename FROM
pictures_archive p inner join pictures_archive_category pc
on p.pictures_archive_id = pc.pictures_archive_id
where pc.pictures_category_id=9
answered Sep 26, 2012 at 8:35
AnandPhadkeAnandPhadke
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So I’ve just started learning MySQL with a few exercises and I’m currently stuck at these 2 problems. Since this is technically an exercise I’d really appreciate hints instead of full solutions. Here are the 2 tables first:
CREATE TABLE LINE (
INV_NUMBER int,
LINE_NUMBER int,
P_CODE varchar(10),
LINE_UNITS float(8),
LINE_PRICE float(8),
LINE_TOTAL float(8)
);
INSERT INTO LINE VALUES('1001','1','13-Q2/P2','1','14.99','14.99');
INSERT INTO LINE VALUES('1001','2','23109-HB','1','9.95','9.95');
INSERT INTO LINE VALUES('1002','1','54778-2T','2','4.99','9.98');
INSERT INTO LINE VALUES('1003','1','2238/QPD','1','38.95','38.95');
INSERT INTO LINE VALUES('1003','2','1546-QQ2','1','39.95','39.95');
INSERT INTO LINE VALUES('1003','3','13-Q2/P2','5','14.99','74.95');
INSERT INTO LINE VALUES('1004','1','54778-2T','3','4.99','14.97');
INSERT INTO LINE VALUES('1004','2','23109-HB','2','9.95','19.90');
INSERT INTO LINE VALUES('1005','1','PVC23DRT','12','5.87','70.44');
INSERT INTO LINE VALUES('1006','1','SM-18277','3','6.99','20.97');
INSERT INTO LINE VALUES('1006','2','2232/QTY','1','109.92','109.92');
INSERT INTO LINE VALUES('1006','3','23109-HB','1','9.95','9.95');
INSERT INTO LINE VALUES('1006','4','89-WRE-Q','1','256.99','256.99');
INSERT INTO LINE VALUES('1007','1','13-Q2/P2','2','14.99','29.98');
INSERT INTO LINE VALUES('1007','2','54778-2T','1','4.99','4.99');
INSERT INTO LINE VALUES('1008','1','PVC23DRT','5','5.87','29.35');
INSERT INTO LINE VALUES('1008','2','WR3/TT3','3','119.95','359.85');
INSERT INTO LINE VALUES('1008','3','23109-HB','1','9.95','9.95');
CREATE TABLE PRODUCT (
P_CODE varchar(10),
P_DESCRIPT varchar(35),
P_INDATE date,
P_QOH int,
P_MIN int,
P_PRICE float(8),
P_DISCOUNT float(8),
V_CODE int
);
INSERT INTO PRODUCT VALUES('11QER/31','Power painter, 15 psi., 3-
nozzle','2015-11-03','8','5','109.99','0','25595');
INSERT INTO PRODUCT VALUES('13-Q2/P2','7.25-in. pwr. saw blade','2015-12-
13','32','15','14.99','0.05',NULL);
INSERT INTO PRODUCT VALUES('14-Q1/L3','9.00-in. pwr. saw blade','2015-11-
13','18','12','17.49','0','21344');
INSERT INTO PRODUCT VALUES('1546-QQ2','Hrd. cloth, 1/4-in., 2x50','2016-01-
15','15','8','39.95','0','23119');
INSERT INTO PRODUCT VALUES('1558-QW1','Hrd. cloth, 1/2-in., 3x50','2016-01-
15','23','5','43.99','0','23119');
INSERT INTO PRODUCT VALUES('2232/QTY','B&D jigsaw, 12-in. blade','2015-12-
30','8','5','109.92','0.05','24288');
INSERT INTO PRODUCT VALUES('2232/QWE','B&D jigsaw, 8-in. blade','2015-12-
24','6','5','99.87','0.05','24288');
INSERT INTO PRODUCT VALUES('2238/QPD','B&D cordless drill, 1/2-in.','2016-
01-20','12','5','38.95','0.05','25595');
INSERT INTO PRODUCT VALUES('23109-HB','Claw hammer','2016-01-
12','23','10','9.95','0.1','21225');
INSERT INTO PRODUCT VALUES('23114-AA','Sledge hammer, 12 lb.','2016-01-
2','8','5','14.40','0.05',NULL);
INSERT INTO PRODUCT VALUES('54778-2T','Rat-tail file, 1/8-in. fine','2015-
12-15','43','20','4.99','0','21344');
INSERT INTO PRODUCT VALUES('89-WRE-Q','Hicut chain saw, 16 in.','2016-02-
17','11','5','256.99','0.05','24288');
INSERT INTO PRODUCT VALUES('PVC23DRT','PVC pipe, 3.5-in., 8-ft','2016-02-
27','188','75','5.87','0','24004');
INSERT INTO PRODUCT VALUES('SM-18277','1.25-in. metal screw, ''25','2016-03-
01','172','75','6.99','0','21225');
INSERT INTO PRODUCT VALUES('SW-23116','2.5-in. wd. screw, 50','2016-02-
14','237','100','8.45','0','21231');
INSERT INTO PRODUCT VALUES('WR3/TT3','Steel matting, 4''x8''x1/6", .5"
mesh','2016-01-27','18','5','119.95','0.1','25595');
PKs are line.INV_NUMBER, line.LINE_NUMBER, product.P_CODE and FKs are line.P_CODE and line.V_CODE
So the first question is to list all product sales that are greater than the average units sold for that product and add a correlated in-line sub-query to list the average units sold per product.
If I understood correctly this means that I need to sum the line.LINE_UNITS group by P_CODE then compare it to the average units sold but I keep getting the same error that says sub query returns more than 1 row. Here’s my code:
select P_CODE,
LINE_UNITS,
( SELECT AVG(line.LINE_UNITS) ) as ‘Unit_Average’
from line
where
( SELECT sum(line.LINE_UNITS)
from line
group by P_CODE ) >
( SELECT AVG(line.LINE_UNITS) );
I think I need to do a join instead and I’d really appreciate some hints.
For the second question I am supposed to list the difference between each product’s prices and the average product price. I managed to type up a few lines but I couldn’t get it to execute for all rows of the product table. I’m thinking of using over() but im having quite a lot of syntax errors. Here’s my code:
select P_CODE, P_PRICE,
( SELECT avg(product.P_PRICE) ) as 'Average',
if(product.P_PRICE <=
( SELECT avg(product.P_PRICE) ),
( SELECT avg(product.P_PRICE) ) - product.P_PRICE,
product.P_PRICE - ( SELECT avg(product.P_PRICE) )
) as 'Difference'
from product;
Thanks guys!
у меня есть две таблицы в БД со следующей структурой:
Таблица 1: 3 строки — category_id, product_id и position
таблица 2: 3 строки — category_id, product_id и position
я пытаюсь установить положение таблицы 1 в положение таблицы 2, где категория и идентификатор продукта совпадают с таблицами.
ниже SQL, я пытался сделать это, но возвращает ошибку MySQL 1242 — подзапрос возвращает более 1 строки
UPDATE table1
SET position = (
SELECT position
FROM table2
WHERE table1.product_id = table2.product_id AND table1.category_id = table2.category_id
)
1
Решение
Решение очень простое и может быть сделано в два простых шага. Первый шаг — это предварительный просмотр того, что будет изменено, чтобы избежать уничтожения данных. Это можно пропустить, если вы уверены в своем WHERE пункт.
Шаг 1: предварительный просмотр изменений
Объедините таблицы, используя поля, которые вы хотите сопоставить, выберите все для визуальной проверки соответствия.
SELECT t1.*, t2.*
FROM table1 t1
INNER JOIN table2 t2
ON t1.category_id = t2.category_id
AND t1.product_id = t2.product_id
Вы также можете добавить WHERE предложение, если только некоторые строки должны быть изменены.
Шаг 2: сделать актуальное обновление
Заменить SELECT пункт и FROM ключевое слово с UPDATE, добавить SET пункт, где это принадлежит. Держать WHERE пункт:
UPDATE table1 t1
INNER JOIN table2 t2
ON t1.category_id = t2.category_id
AND t1.product_id = t2.product_id
SET t1.position = t2.position
Это все.
Технические соображения
Индексы столбцов, используемых на JOIN предложение в обеих таблицах является обязательным, если в таблицах более нескольких сотен строк. Если запрос не имеет WHERE условия, то MySQL будет использовать индексы только для самой большой таблицы. Индексы полей, используемых на WHERE условие ускорит запрос. Prepend EXPLAIN к SELECT запрос, чтобы проверить план выполнения и решить, какие индексы вам нужны.
Можете добавить SORT BY а также LIMIT для дальнейшего сокращения набора измененных строк, используя критерии, которые не могут быть достигнуты с помощью WHERE (например, только самые последние / самые старые 100 строк и т. д.). Поместите их на SELECT сначала запрос, чтобы проверить результат, а затем изменить SELECT в UPDATE как описано.
Конечно, индексы на столбцах, используемых на SORT BY пункт является обязательным.
1
Другие решения
Вы можете запустить этот запрос, чтобы увидеть, что происходит:
SELECT product_id, category_id, count(*), min(position), max(position)
FROM table2
GROUP BY product_id, category_id
HAVING COUNT(*) > 1;
Это даст вам список product_id, category_id пары, которые появляются несколько раз в table2, Тогда вы можете решить, что делать. Вы хотите произвольное значение position? Является ли значение position всегда одно и то же? Вам нужно починить стол?
Достаточно легко решить конкретную проблему с помощью limit 1 или функция агрегации. Тем не менее, вам может понадобиться исправить данные в таблице. Исправление выглядит так:
UPDATE table1 t1
SET t1.position = (SELECT t2.position
FROM table2 t2
WHERE t2.product_id = t1.product_id AND t2.category_id = t1.category_id
LIMIT 1
);
0