I am relatively new to SQL and I am trying to extract rows where they have the highest values.
For example, the table look like this:
user_id fruits
1 apple
1 orange
2 apple
1 pear
I would like to extract the data such that it would look like this:
user_id fruits
1 3
If user_id 2 has 3 fruits, it should display:
user_id fruits
1 3
2 3
I can only manage to get the if I use LIMIT = 1 by DESC order, but that is not the right way to do it. Otherwise I am getting only:
user_id fruits
1 3
2 1
Not sure where to store the max value to put in the where clause. Appreciate any help, thank you
Use RANK():
WITH cte AS (
SELECT user_id, COUNT(*) AS cnt, RANK() OVER (ORDER BY COUNT(*) DESC) rnk
FROM yourTable
GROUP BY user_id
)
SELECT user_id, cnt AS fruits
FROM cte
WHERE rnk = 1;
Here's one answer (with sample data):
CREATE TABLE something (user_id INT NOT NULL, fruits VARCHAR(10) NOT NULL, PRIMARY KEY (user_id, fruits));
INSERT INTO something VALUES (1, 'apple');
INSERT INTO something VALUES (1, 'orange');
INSERT INTO something VALUES (2, 'apple');
INSERT INTO something VALUES (1, 'pear');
INSERT INTO something VALUES (2, 'orange');
INSERT INTO something VALUES (2, 'pear');
SELECT user_id, COUNT(*) AS cnt
FROM something
GROUP BY user_id
HAVING COUNT(*) >= ALL (SELECT COUNT(*) FROM something GROUP BY user_id);
Related
I have two tables:
TABLE A
Unique_id
id
price
1
1
10.50
2
3
14.70
3
1
12.44
TABLE B
Unique_id
Date
Category
Store
Cost
1
2022/03/12
Shoes
A
13.24
2
2022/04/15
Hats
A
15.24
3
2021/11/03
Shoes
B
22.31
4
2000/12/14
Shoes
A
15.33
I need to filter TABLE A on a known id to get the Unique_id and average price to join to Table B.
Using this information I need to know which stores this item was sold in.
I then need to create a results table displaying the stores and the amount of days sales were recorded in the stores - regardless of whether the sales are associated with the id and the average cost.
To put it more simply I can break down the task into 2 separate commands:
SELECT AVG(price)
FROM table_a
WHERE id = 1
GROUP BY unique_id;
SELECT store, COUNT(date), AVG(cost)
FROM table_b
WHERE category = 'Shoes'
GROUP BY store;
The unique_id should inform the join but when I join the tables it messes up my COUNT function and only counts the days in which the id is connected - not the total store sales days.
The results should look something like this:
Store
AVG price
COUNT days
AVG cost
A
10.50.
3
14.60.
B
12.44
1.
22.31.
I wwas hard to grasp, what you wanted, but after some thinking and your clarification, it can be solved as the code shows
CREATE TABLE TableA
(`Unique_id` int, `id` int, `price` DECIMAL(10,2))
;
INSERT INTO TableA
(`Unique_id`, `id`, `price`)
VALUES
(1, 1, 10.50),
(2, 3, 14.70),
(3, 1, 12.44)
;
CREATE TABLE TableB
(`Unique_id` int, `Date` datetime, `Category` varchar(5), `Store` varchar(1), `Cost` DECIMAL(10,2))
;
INSERT INTO TableB
(`Unique_id`, `Date`, `Category`, `Store`, `Cost`)
VALUES
(1, '2022-03-12 01:00:00', 'Shoes', 'A', 13.24),
(2, '2022-04-15 02:00:00', 'Hats', 'A', 15.24),
(3, '2021-11-03 01:00:00', 'Shoes', 'B', 22.31),
(4, '2000-12-14 01:00:00', 'Shoes', 'A', 15.33)
SELECT
B.`Store`
, AVG(A.`price`) price
, (SELECT COUNT(*) FROM TableB WHERE `Store` = B.`Store` ) count_
, (SELECT AVG(
`cost`) FROM TableB WHERE `Store` = B.`Store` ) price
FROM TableA A
JOIN TableB B ON A.`Unique_id` = B.`Unique_id`
WHERE B.`Category` = 'Shoes'
GROUP BY B.`Store`
Store | price | count_ | price
:---- | --------: | -----: | --------:
A | 10.500000 | 3 | 14.603333
B | 12.440000 | 1 | 22.310000
db<>fiddle here
This should be the query you are after. Mainly you simply join the rows using an outer join, because not every table_b row has a match in table_a.
Then, the only hindrance is that you only want to consider shoes in your average price. For this to happen you use conditional aggregation (a CASE expression inside the aggregation function).
select
b.store,
avg(case when b.category = 'Shoes' then a.price end) as avg_shoe_price,
count(b.unique_id) as count_b_rows,
avg(b.cost) as avg_cost
from table_b b
left outer join table_a a on a.unique_id = b.unique_id
group by b.store
order by b.store;
I must admit, it took me ages to understand what you want and where these numbers result from. The main reason for this is that you have WHERE table_a.id = 1 in your query, but this must not be applied to get the result you are showing. Next time please look to it that your description, queries and sample data match.
(And then, I think that names like table_a, table_b and unique_id don't help understanding this. If table_a were called prices instead and table_b costs and unique_id were called cost_id then, I wouldn't have had to wonder how the tables are related (by id? by unique id?) and wouldn't have had to look again and again which table the cost resides in, which table has a price and which table is the outer joined one while looking at the problem, the requested result and while writing my query.)
my table:
drop table if exists new_table;
create table if not exists new_table(
obj_type int(4),
user_id varchar(30),
payer_id varchar(30)
);
insert into new_table (obj_type, user_id, payer_id) values
(1, 'user1', 'payer1'),
(1, 'user2', 'payer1'),
(2, 'user3', 'payer1'),
(1, 'user1', 'payer2'),
(1, 'user2', 'payer2'),
(2, 'user3', 'payer2'),
(3, 'user1', 'payer3'),
(3, 'user2', 'payer3');
I am trying to select all the payer id's whose obj_type is only one value and not any other values. In other words, even though each payer has multiple users, I only want the payers who are only using one obj_type.
I have tried using a query like this:
select * from new_table
where obj_type = 1
group by payer_id;
But this returns rows whose payers also have other user's with other obj_types. I am trying to get a result that looks like:
obj | user | payer
----|-------|--------
3 | user1 | payer3
3 | user2 | payer3
Thanks in advance.
That is actually easy:
SELECT player_id
FROM new_table
GROUP BY player_id
HAVING COUNT(DISTINCT obj_type) = 1
Having filters rows just like WHERE but it does so after the aggregation.
The difference is best explained by an example:
SELECT dept_id, SUM(salary)
FROM employees
WHERE salary > 100000
GROUP BY dept_id
This will give you the sum of the salaries of people earning more than 100000 each.
SELECT dept_id, SUM(salary)
FROM employees
GROUP BY dept_id
HAVINF salary > 100000
The second query will give you the departments where all employees together earn more than 100000 even if no single employee earns that much.
If you want to return all rows without grouping them you can use analytic functions:
SELECT * FROM (
SELECT obj_type,user_id,
payer_id,
COUNT(DISTINCT obj_type) OVER (PARTITION BY payer_id) AS distinct_obj_type
FROM new_table)
WHERE distinct_obj_type = 1
Or you can use exist with the query above:
SELECT *
FROM new_table
WHERE payer_id IN (SELECT payer_id
FROM new_table
GROUP BY payer_id
HAVING COUNT(DISTINCT obj_type) = 1)
I have table like this
mapname id time
---------------------
cvjourney 1 65.24
cvjourney 3 69.45
cvjourney 2 64.25
cvjourney 8 75.45
dark_hop 2 65.87
dark_hop 1 61.51
coldrunss 1 12.50
I want to select data from that table to get something like this (for id 1, 2)
cvjourney 1 65.24
cvjourney 2 64.25
dark_hop 2 65.87
dark_hop 1 61.51
so I want select data where two players are both having records on same map, I dont want select maps where player[1] has record and player[2] doesnt, or the other way.
Here is my select, but I cant get rid of maps where one player has record and another doesnt.
SELECT *
FROM table
WHERE mapname IN (
SELECT mapname
FROM table
WHERE id IN ('1','2')
)
AND id IN ('1', '2')
ORDER BY mapname
Can you help me, please?
You can use aggregation to get only those mapnames which have both the ids present (which means after applying the id filter, distinct count of ids will be two) and then use that to get the relevant rows.
select t1.*
from table t1
join (
select mapname
from table
where id in (1, 2)
group by mapname
having count(distinct id) = 2
) t2 on t1.mapname = t2.mapname
where t1.id in (1, 2)
order by t1.mapname
Use can use IN too:
select t1.*
from table t1
where mapname in (
select mapname
from table
where id in (1, 2)
group by mapname
having count(distinct id) = 2
)
and id in (1, 2)
order by mapname
Demo
I ran into a problem trying to pull one action per user with the least priority, the priority is based on other columns content and is an integer,
This is the initial query :
SELECT
CASE
...
END AS dummy_priority,
id,
user_id
FROM
actions
Result :
id user_id priority
1 2345 1
2 2345 3
3 2999 5
4 2999 2
5 3000 10
Desired result :
id user_id priority
1 2345 1
4 2999 2
5 3000 10
Following what i want i tried
SELECT x.id, x.user_id, MIN(x.priority)
FROM (
SELECT
CASE
...
END AS priority,
id,
user_id
FROM
actions
) x
GROUP BY x.user_id
Which didn't work
Error Code: 1055. Expression #1 of SELECT list is not in GROUP BY
clause and contains nonaggregated column 'x.id' which is not
functionally dependent on columns in GROUP BY clause;
Most examples of this I found were extracting just the user_id and priority and then doing an inner join with both of them to get the row, but I can't do that since (priority, user_id) isn't unique
A simple verifiable example would be
CREATE TABLE `actions` (
`id` int(11) NOT NULL,
`user_id` int(11) DEFAULT NULL,
`priority` int(11) DEFAULT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
INSERT INTO `actions` (`id`, `user_id`, `priority`) VALUES
(1, 2345, 1),
(2, 2345, 3),
(3, 2999, 5),
(4, 2999, 2),
(5, 3000, 10);
how to extract the desired result (please hold in mind that this table is a subquery)?
The proper way to do this would involve a subquery of some sort . . . and that would require repeating the case definition.
Here is another method, using the substring_index()/group_concat() trick:
SELECT SUBSTRING_INDEX(GROUP_CONCAT(x.id ORDER BY x.priority), ',', 1) as id,
x.user_id, MIN(x.priority)
FROM (SELECT (CASE ...
END) AS priority,
id, user_id
FROM actions a
) x
GROUP BY x.user_id;
And that proper way in full...
SELECT x...
, CASE...x... priority
FROM my_table x
JOIN
( SELECT user_id
, MIN(CASE...) priority
FROM my_table
GROUP
BY user_id
) y
ON y.user_id = x.user_id
AND y.priority = CASE...x...;
This should work ...
SELECT id , user_id, priority FROM actions act
INNER JOIN
(SELECT
user_id, MIN(priority) AS priority
FROM
actions
GROUP BY user_id) pri
ON act.user_id = pri.user_id AND act.priority = pri.prority
Is it possible to select distinct company names from the customer table but also displaying the iD's related?
at the minute I'm using
SELECT company,id, COUNT(*) as count FROM customers GROUP BY company HAVING COUNT(*) > 1;
which returns
MyDuplicateCompany1 64 2
MyDuplicateCompany2 20 3
MyDuplicateCompany6 175 2
but what I'm after is all the duplicate ID's for each.
so
CompanyName, TimesDuplicated, DuplicateId1, DuplicateId2, DuplicateId3
or a row for each so
MyDuplicateCompany1, DuplicateId1, TimesDuplicated
MyDuplicateCompany1, DuplicateId2, TimesDuplicated
MyDuplicateCompany2, DuplicateId1, TimesDuplicated
MyDuplicateCompany2, DuplicateId2, TimesDuplicated
MyDuplicateCompany2, DuplicateId3, TimesDuplicated
is this possible?
Not sure if this would be acceptable but there's a function in mySQL which allows you to combine multiple rows into one Group_Concat(Field), but show the distinct values for each record for columns specified (like ID in this case)
SELECT company
, COUNT(*) as count
, group_concat(ID) as DupCompanyIDs
FROM customers
GROUP BY company
HAVING COUNT(*) > 1;
SQL Fiddle
showing similar results with duplicate companies listed in one field.
If you need it in multiple columns or multiple rows, you could wrap the above as an inline view and inner join it back to customers on the name to list the duplicates and times duplicated.
You can use GROUP_CONCAT(id) to concat your id by comma, your query should be:
SELECT company, GROUP_CONCAT(id) as ids, COUNT(id) as cant FROM customers GROUP BY company HAVING cant > 1
You can test the query with this
CREATE TABLE IF NOT EXISTS `customers` (
`id` int(11) NOT NULL,
`company` varchar(50) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
INSERT INTO `customers` (`id`, `company`) VALUES
(1, 'MyDuplicateCompany1'),
(2, 'MyDuplicateCompany1'),
(3, 'MyDuplicateCompany1'),
(4, 'MyDuplicateCompany2'),
(5, 'MyDuplicateCompany2'),
(6, 'MyDuplicateCompany3'),
(7, 'MyDuplicateCompany3'),
(8, 'MyDuplicateCompany3'),
(9, 'MyDuplicateCompany3'),
(10, 'MyDuplicateCompany4');
Output:
Read more at:
http://monksealsoftware.com/mysql-group_concat-and-postgres-array_agg/
You are not looking for companies with more than 1 entry (GROUP BY company), but for duplicate company IDs (GROUP BY company, id):
SELECT company, id, COUNT(*)
FROM customers
GROUP BY company, id
HAVING COUNT(*) > 1;
This should give exactly what you're looking for without GROUP_CONCAT()
SELECT
company, id,
( SELECT COUNT(*) from customers AS b
WHERE a.company = b.company
) AS cnt
FROM customers AS a
GROUP BY company, id
HAVING cnt > 1
;
Note: GROUP_CONCAT does the same thing, just all in one row per company.