Get min price id without inner select - mysql

I have a table called a with this data:
+-----+-----------+-------+
| id | parent_id | price |
+-----+-----------+-------+
| 1 | 1 | 100 |
| 2 | 1 | 200 |
| 3 | 1 | 99 |
| 4 | 2 | 1000 |
| 5 | 2 | 999 |
+-----+-----------+-------+
I want to get the id of min pirce for each parent_id.
There is any way to get this result without subquery?
+-----+-----------+-------+
| id | parent_id | price |
+-----+-----------+-------+
| 3 | 1 | 99 |
| 5 | 2 | 999 |
+-----+-----------+-------+


SELECT D1.id, D1.parent_id, D1.price
FROM Data D1
LEFT JOIN Data D2 on D2.price < D1.price AND D1.parent_id = D2.parent_id
WHERE D2.id IS NULL


Here is a shot at how to do it without subqueries. I haven't tested, let me know if it works!
SELECT t.id, t.parent_id, t.price
FROM table t
LEFT JOIN table t2
ON (t.parent_id = t2.parent_id AND t.price > t2.price)
GROUP BY t.id, t.parent_id, t.price
HAVING COUNT(*) = 1 AND max(t2.price) is null
ORDER BY t.parent_id, t.price desc;


Try this:
SELECT T1.id,T2.parent_id,T2.price FROM
(SELECT id,price
FROM TableName) T1
INNER JOIN
(
SELECT parent_id,MIN(price) as price
FROM TableName
GROUP BY parent_id) T2 ON T1.price=T2.price
See result in SQL Fiddle.


Try group by,
SELECT parent_id,min(price)
FROM TableName
GROUP BY parent_id


You can do this with a LEFT JOIN
SELECT a.id, a.parent_id, a.price
FROM a
LEFT JOIN a AS b ON b.price < a.price AND b.parent_id = a.parent_id
WHERE b.id IS NULL
Find the results at this fiddle:
http://sqlfiddle.com/#!2/09c888/10


You can try this without using any join or subquery you will surely get the desired result.
SELECT TOP 2 FROM a ORDER BY price

Related

Display child and parent relationship (if any) in a same table

I have this table
| id |parent|name|
| 1 | NULL | E |
| 2 | NULL | B |
| 3 | 5 | U |
| 4 | 5 | X |
| 5 | NULL | C |
| 6 | NULL | A |
I would like the list, ordered by parent's name, of all ID whether they have a parent or not:
| id |parent|name|has_child|
| 6 | NULL | A | 0 |
| 2 | NULL | B | 0 |
| 5 | NULL | C | 1 |
| 3 | 5 | U | 0 |
| 4 | 5 | X | 0 |
| 1 | NULL | E | 0 |
Is it possible?
I have tried many things but never get the proper answer, and I don't really know how to add the 'has_child' column
SELECT
t1.parent,
t2.name
FROM tablename AS t1
INNER JOIN
(
SELECT MIN(id) AS id, parent
FROM tablename
GROUP BY parent
) AS t22 ON t22.id = t1.id AND t1.parent = t22.parent
INNER JOIN tablename AS t2 ON t1.parent = t2.id;

I would use a self join here:
SELECT DISTINCT
t1.id,
t1.parent,
t1.name,
1 - ISNULL(t2.id) has_child
FROM tablename t1
LEFT JOIN tablename t2
ON t1.id = t2.parent
ORDER BY
t1.id;
The join condition used here, which matches a given record as a parent to one or more children, is that the current id is also the parent of some other record(s). Note that we need SELECT DISTINCT here, because a given parent might match to more than one child record.

You can use a self join -- because you want the name of the parent and not the id -- and coalesce() for ordering:
select t.*,
(case when exists (select 1 from t tc where tc.parent = t.id)
then 1 else 0
end)
from t left join
t tp
on t.parent = tp.id
order by coalesce(tp.name, t.name), -- group rows by the parent, if any
(tp.name is null) desc, -- put parent first
t.name; -- order by children

I hope that you find this answer a little bit useful. The subquery gets the distinct id of parents and excludes the blanked fills.
SELECT *,
CASE WHEN id IN (SELECT DISTINCT parent
FROM tablename
WHERE parent IS NOT NULL)
THEN '1' ELSE '0'
END AS has_child
FROM tablename
ORDER BY name;

SELECT t1.id, t1.parent, t1.name, MAX(t2.parent is not null) has_child
FROM table t1
LEFT JOIN table t2 ON t1.id = t2.parent_id
GROUP BY t1.id, t1.parent, t1.name

SQL select from 1 x N where all bigger than

I have tables books and bookType which pose a 1 X n relationship.
books
+-----+------------------+----------+-------+
| id | title | bookType | price |
+-----+------------------+----------+-------+
| 1 | Wizard of Oz | 3 | 14 |
| 2 | Huckleberry Finn | 1 | 16 |
| 3 | Harry Potter | 2 | 25 |
| 4 | Moby Dick | 2 | 11 |
+-----+------------------+----------+-------+
bookTypes
+-----+----------+
| id | name |
+-----+----------+
| 1 | Fiction |
| 2 | Drama |
| 3 | Children |
+-----+----------+
How would I retrieve bookTypes where all books are more expensive than e.g. 12($)?
In this case, the expected output would be:
+-----+----------+
| id | name |
+-----+----------+
| 1 | Fiction |
| 3 | Children |
+-----+----------+

You can use not exists:
select t.*
from bookTypes t
where not exists (
select 1
from books b
where b.bookType = t.id and b.price < 12
)
If you want to select book types that also have at least one associated book:
select t.*
from bookTypes t
where
exists (select 1 from books b where b.bookType = t.id)
and not exists (select 1 from books b where b.bookType = t.id and b.price < 12)

Do a GROUP BY, use HAVING to return only booktypes having the lowest price > 12.
SELECT bt.name
FROM bookTypes bt
INNER JOIN books b ON b.bookType = bt.id
group by bt.name
HAVING SUM(b.price <= 12) = 0;

You can directly consider using having min(price) >= 12 with grouping by bookType
select t.id, t.name
from bookTypes t
join books b
on t.id = b.bookType
group by b.bookType
having min(price) >= 12
Moreover, if your DB's version is at least 10.2, then you can also use some window functions for analytical queries such as min(..) over (partition by .. order by ..) :
with t as
(
select t.id, t.name, min(price) over (partition by bookType) as price
from bookTypes t
join books b
on t.id = b.bookType
)
select id, name
from t
where price >= 12
in which min() over (..) window function determines minimum price for each booktype by use of partition by bookType
Demo

I think GMB's solution is likely the best so far. But for sake of completeness: You can also use the ALL operator with a correlated subquery. That's probably the most straight forward solution.
SELECT *
FROM booktypes bt
WHERE 12 < ALL (SELECT b.price
FROM books b
WHERE b.booktype = bt.id);

Can you not just select from books inner join bookTypes on id WHERE price > 12?
SELECT bt.*
FROM bookTypes bt
INNER JOIN books b ON b.bookType = bt.id
WHERE b.price > 12

Is there any way to do calculate the MAX() quickly?

I am trying to do this query:
SELECT
A.*
, (SELECT MAX(B.Date2) FROM Tab2 B WHERE A.ID = B.ID AND A.Date > B.Date2) AS MaxDate
FROM
Tab A
This works but it takes a lot of time to run when you have a lot of rows. Is there any quicker way to do this which give the same results?
Thank you!
Edit:
The table définitions are as follow:
Tab : (dd-mm-yyyy)
ID | Date
1 | 19-01-2018
1 | 14-01-2018
2 | 18-02-2019
3 | 20-03-2019
Tab2:
ID | Date2
1 | 10-01-2018
1 | 15-01-2018
1 | 20-01-2018
2 | 15-02-2019
2 | 21-02-2019
3 | 25-03-2019
I want my query returns:
ID | Date | MaxDate
1 | 19-01-2018 | 15-01-2018
1 | 14-01-2018 | 10-01-2018
2 | 18-02-2019 | 15-02-2019
3 | 20-03-2019 | NULL
Thanks!

It was unexpected for me but this query worked:
SELECT
A.ID
, A.Date
, MAX(B.Date2) AS MaxDate
FROM
Tab A
left outer join Tab2 B
on A.ID = B.ID and A.Date > B.Date2
GROUP BY
A.ID, A.Date
;
I didn't know that we can put a column from a table in a group by when the column of the MAX() is in another table.

How to join and give a default if a value is in one table but not another?

I am using MySQL, and I'm a newbie!
Hope you guys here can help me with a SQL question.
Say I have 2 tables, and I want to a simple join.
Table 1:
id | service_id | user_number
----------------------------------------------------------------------
0 | 1001 | 10
1 | 1002 | 20
2 | 1004 | 40
Table 2:
id | service_id | error_number
----------------------------------------------------------------------
0 | 1001 | 1000
1 | 1003 | 3000
2 | 1004 | 4000
I want to do a join on service_id and have default value of user_number and error_number to be 0 if it does not exist.
So:
id | service_id | user_number | error_number
----------------------------------------------------------------------
0 | 1001 | 10 | 1000
1 | 1002 | 20 | 0
3 | 1003 | 0 | 3000
2 | 1004 | 40 | 4000
I tried some queries, but they kept giving me null instead of 0.
Thanks a lot.

Here you should use union first, then do aggregation:
select t.`service_id`, sum(t.`user_number`) as `user_number`, sum(t.`error_number`) as `error_number`
from (
select `service_id`, `user_number`, 0 as `error_number` from t1
union
select `service_id`, 0 as `user_number`, `error_number` from t2
) t
group by `service_id`
demo here.

You can try this one, mate:
SELECT
t1.id,
t1.service_id,
COALESCE(tb1.user_number, 0) `user_number`,
COALESCE(tb2.error_number, 0) `error_number`
FROM
(
SELECT id, service_id
FROM table1
UNION
SELECT id, service_id
FROM table2
) t1
LEFT JOIN table1 tb1 ON tb1.service_id = t1.service_id
LEFT JOIN table2 tb2 ON tb2.service_id = t1.service_id;

Try this:
select COALESCE(t1_service,t2_service ) as service_id, COALESCE(user_number,0) as user_number , COALESCE(error_number,0) as error_number
from (
select t1.service_id as t1_service , t1.user_number , t2.error_number, t2.service_id as t2_service
from table_1 t1
LEFT OUTER JOIN table_2 t2
on t1.service = t2.service
union
select t1.service_id as t1_service , t1.user_number , t2.error_number, t2.service_id as t2_service
from table_1 t1
Right OUTER JOIN table_2 t2
on t1.service = t2.service
)z1
order by service_id

Select last record of each person

I have the following tables
tbl_investors
id | first_name | last_name |
---------------------------------------
1 | Jon | Cold |
2 | Rob | Ark |
3 | Rickon | Bolt |
tbl_investors_ledger
id | investor_id | amount |
------------------------------------
1 | 1 | 500 |
2 | 2 | 200 |
3 | 2 | 250 |
4 | 2 | 300 |
5 | 3 | 10 |
6 | 1 | 550 |
7 | 3 | 20 |
I just want to return all investors with their latest amount. Ex, Jon Cold with 550, Rob Ark 300 and Rickon Bolt 20, alphabetically with their last name.
I have an existing query but it will not return the latest amount of the investor. Can someone help me what i'm doing wrong?
SELECT t1.*, t2.*
FROM ".tbl_investors." t1
LEFT JOIN ".tbl_investors_ledger." t2
ON t1.id = t2.investor_id
LEFT JOIN (SELECT t.investor_id, max(t.id) as tid
FROM ".tbl_investors_ledger." t ) tt
ON tt.investor_id = t2.investor_id AND tt.tid = t2.id
GROUP BY t2.investor_id
ORDER BY t1.last_name

You can use GROUP_CONCAT and SUBSTRING_INDEX together
SELECT I.*
, SUBSTRING_INDEX(GROUP_CONCAT(L.amount ORDER BY L.id DESC), ',', 1) AS LastAmount
FROM tbl_investors AS I
LEFT JOIN tbl_investors_ledgers AS L
ON L.investor_id = I.id
GROUP BY I.id
ORDER BY I.last_name
Here a demo from SQLFiddle, many thanks to #zakhefron :)

Try this;)
SELECT t1.*, t2.*
FROM tbl_investors t1
LEFT JOIN tbl_investors_ledger t2
ON t1.id = t2.investor_id
INNER JOIN (
SELECT t.investor_id, max(t.id) as tid
FROM tbl_investors_ledger t GROUP BY t.investor_id) tt
ON tt.investor_id = t2.investor_id AND tt.tid = t2.id
ORDER BY t1.last_name
SQLFiddle DEMO
And check related OP Retrieving the last record in each group and this blog How to select the first/least/max row per group in SQL, you can find more solutions for your question.