Was posted yesterday at:
MySQL: Finding most frequently occuring values in table
But was wondering if there is a way to refine the answer, as you shouldn't have to repeat the code at the end to get the MAX(COUNT)
Question
I have two tables:
Purchases:
item
001
003
002
001
002
004
003
001
002
Item:
id | name
001 | Item 1
002 | Item 2
003 | Item 3
004 | Item 4
Expected output:
item name
001 Item 1
002 Item 2
I need to find the (multiple) items that occur most frequently in the purchases table, and output the name of the item. What should I have in my query?
Answer I have so far
SELECT t.cnt, t.name FROM
(SELECT COUNT(*) AS "cnt", item.name
FROM purchases
LEFT JOIN item ON item.id = purchases.item
GROUP BY item.name) t
WHERE t.cnt = (SELECT MAX(t2.cnt) FROM
(SELECT COUNT(*) AS "cnt"
FROM purchases
LEFT JOIN item ON item.id = purchases.item
GROUP BY item.name) t2
)
You must first obtain the maximal count, then use that to filter for matching items:
SELECT i.*
FROM purchases p
JOIN item i ON i.id = p.item
GROUP BY i.id
HAVING COUNT(*) = (
SELECT COUNT(*)
FROM purchases
GROUP BY item
ORDER BY COUNT(*) DESC
LIMIT 1
)
Note that this query relies on id being the PK of (or at very least unique within) your item table, for two reasons:
It ensures that the JOIN does not adversely affect the COUNT(*) in the outer query; and
It enables you to determinately select * from item despite the grouping operation (other RDBMS platforms wouldn't allow this, but would require you to group on every non-aggregated output column—but MySQL offers this "feature" as a performance improvement).
I believe something like this could work. You can using 'having' and 'order by' to get top results.
SELECT count(p.item), i.name
FROM purchases p, item i
WHERE p.item = i.id
GROUP BY p.item HAVING count(p.item) > 1
ORDER BY count(p.item) DESC;
Optionally you could add 'LIMIT 1' to the end.
Try this:
SELECT I.`id` AS id, COUNT(I.`id`) AS `id_count`, P.`name` AS name
FROM `item` I
JOIN `purchases` P
ON I.`id` = P.`id`
ORDER BY COUNT(I.`id`) DESC
GROUP BY I.`id`
Expected Output:
|-------|-------------|--------------------|
| id | id_count | name |
|-------|-------------|-------------------|
| 101 | 5 | Item 1 |
| 102 | 6 | Item 2 |
| 103 | 8 | Item 3 |
| 104 | 12 | Item 4 |
...
|------------------------------------------|
Related
I have two tables
tbl_groups:
id | name
----------------
1 | BSCS
2 | BSIT
3 | BBA
tbl_students:
id | name | group_id
-------------------------------
1 | Student Name | 1
2 | Student 2 | 1
3 | Student 3 | 2
I want to show groups details: group name and number of students in a particular group,
I am using this query but it shows groups that has students. it does not show group with 0 students.
select tb2.id, tb2.name, count(*) from tbl_students tb1 JOIN tbl_groups tb2 ON tb1.group_id = tb2.id
How do I show all groups, please give me some idea
EDIT:
if I use above query I get following result:
id | name | count(*)
-------------------------------
1 | Student Name | 2
2 | BSIT | 1
(it doest show 3rd group because there are 0 students, I want to show this groups also).
Just use a left join:
select tb2.id, tb2.name, count(tb1.id) as no_std
from tbl_groups tb2
LEFT JOIN tbl_students tb1 ON tb2.id = tb1.group_id
group by tb2.id, tb2.name
See it working live here: http://sqlfiddle.com/#!9/2282a3/5
I would just use a correlated subquery to get the count of students in each group, like so:
select
g.*,
(select count(*) from tbl_students s where s.group_id = g.id) no_students
from tbl_groups g
This does not filter out groups that have no students (it will give a count of 0 instead). And with an index on tbl_students(group_id), this should be as efficient as it gets (this index is already there if you set up a foreign key constraint on that column - as you should have).
I have a table like this (simplified version):
+------+-------+-----+--------------+-----+
| id | name | age | company.name | ...
+------+-------+-----+--------------------+
| 1 | Adam | 21 | Google | ...
| 3 | Peter | 20 | Apple | ...
| 2 | Bob | 20 | Microsoft | ...
| 9 | Alice | 18 | Google | ...
+------+-------+-----+--------------------+
I need groups data with counting rows by any one column. And I need to get first row in each group. User select which column will be used to group.
If user select column age to group then results:
+------+------------+-------+
| id | group_name | count |
+------+------------+-------+
| 9 | 18 | 1 |
+------+------------+-------+
| 2 | 20 | 2 |
+------+------------+-------+
| 1 | 21 | 1 |
+------+------------+-------+
Column to group may be numeric or string.
Currently I does it by this query:
SELECT id, group_name, users_name, count(id) as count FROM (
SELECT persons.id as id, company.type as group_name, users.name as users_name
FROM persons
LEFT JOIN company on company.id = persons.company_id
LEFT JOIN position on position.id=persons.position_id
...
LEFT JOIN source on source.id=persons.source_id
WHERE ...
ORDER BY if(company.type = '' or company.type is null,1,0) ASC,
company.type ASC, IF(persons.status = '' or persons.status is null,1,0) ASC,
persons.status ASC, persons.id
) t1 GROUP BY group_name
but with new version mysql this SQL stoped works I think that order is ignored in sub-select.
I know that similar topics was wroted, but proposed solutions not working with my query. I have to join many tables, add multiple conditions and use cascade order and then select first row from each group. I will be very happy if solution will be optimised for performace.
---- EDIT ----
Proposed solution:
SQL select only rows with max value on a column
which suggest to use MAX() and GROUP BY not working well. For two reason
If grouped column include string, then query return not first row, but last row in each group.
If my dataset has a cascade order, I can not use MAX in a few columns at the same time.
I created sqlfiddle which include exact example.
http://sqlfiddle.com/#!9/23225d/11/0
-- EXAMPLE 1 - Group by string
-- base query
SELECT persons.*, company.* FROM persons
LEFT JOIN company ON persons.company_id = company.id
ORDER BY company.name ASC, company.id ASC;
-- grouping query
SELECT MAX(persons.id) as id, company.name, count(persons.id) as count
FROM persons
LEFT JOIN company ON persons.company_id = company.id
GROUP BY company.name
ORDER BY company.name ASC, persons.id ASC;
-- The results will be:
-- |ID | NAME | COUNT|
-- |1 | Google | 2 |
-- |3 | Microsoft| 3 |
-- EXAMPLE 2 - Cascade order
-- base query
SELECT persons.*, company.* FROM persons
LEFT JOIN company ON persons.company_id = company.id
ORDER BY company.type ASC, persons.status ASC;
-- grouping query
SELECT MAX(persons.id) as id, company.type, count(persons.id) as count
FROM persons
LEFT JOIN company ON persons.company_id = company.id
GROUP BY company.type
ORDER BY company.type ASC, persons.status ASC;
-- The results will be:
-- |ID | NAME| COUNT|
-- |3 | 1 | 2 |
-- |2 | 2 | 3 |
Just change MAX() to MIN() to get the first row instead of the last row in each group.
To get the extreme values of cascading columns, see SQL : Using GROUP BY and MAX on multiple columns. Use that in the subquery part of the query to get the row containing those extremes, as in SQL select only rows with max value on a column.
So the form of the full query is:
SELECT t1.id, t1.grouped_column, t2.count
FROM yourTable AS t
JOIN (SELECT t3.grouped_column, t3.order_column1, MIN(t4.order_column2) AS order_column2, SUM(t3.count) AS count
FROM (SELECT grouped_column, MIN(order_column1) AS order_column1, COUNT(*) AS count
FROM yourTable
GROUP BY grouped_column) AS t3
JOIN yourTable AS t4
ON t3.grouped_column = t4.grouped_column AND t3.order_column1 = t4.order_column1
GROUP BY t4.grouped_column, t4.order_column1) AS t2
ON t1.grouped_column = t2.grouped_column AND t1.ordered_column1 = t2.order_column1 AND t1.order_column2 = t2.order_column2
Since you want to operate on a join, I suggest you define a view that uses the join. Then you can use that view in place of yourTable in the above query.
Why this query wont work? Is it beacause combinaton of order by and group by?
One table is with adverts, other with subscriptions, third is with services, and fourth is many to many relation between services and locations (location is position where advert should be shown).
What i want is to order adverts stored in adverts table having location 2 first, then those who don't have location defined and then with location 1 (this order is generated programmicaly)
adverts table:
id, name, subscription_id
subscriptions table:
subscription_id, service_id, date, paid etc...
service_locations table:
service_id, location_id
as you can se there is fourth table in this case, but it is unimportant
The query:
select adverts.id, GROUP_CONCAT(service_locations.location_id) AS locations from adverts
left join subscriptions
on adverts.subscription_id = subscriptions.id
left join service_locations
on service_locations.service_id = subscriptions.service_id
group by adverts.id
order by case service_locations.location_id
when 2 then 1
when 1 then 3
else 2
end
Expected results:
+----+-----------+
| id | locations |
+----+-----------+
| 1 | 2 |
| 3 | 1,2 |
| 2 | null |
+----+-----------+
What i actually get (the third in row has location 2 but it is placed after null):
+----+-----------+
| id | locations |
+----+-----------+
| 1 | 2 |
| 2 | null |
| 3 | 1,2 |
+----+-----------+
When you use group by, all columns not in the group by should have aggregation functions. So, I think you intend something like this:
select a.id, GROUP_CONCAT(sl.location_id) AS locations
from adverts a left join
subscriptions s
on a.subscription_id = s.id left join
service_locations sl
on sl.service_id = s.service_id
group by a.id
order by max(case sl.location_id
when 2 then 1
when 1 then 3
else 2
end);
I'm not sure if max() is what you really need, but you do need an aggregation function. This specifically produces the output in the question:
order by (case min(sl.location_id)
when 2 then 1
when 1 then 2
else 3
end);
I have found a solution, order by must be executed before group by, which is not a default behaivor, more about that behaivour here: https://stackoverflow.com/a/14771322/4329156) (a subquery must be used)
So, query should look like
select *, GROUP_CONCAT(location_id) as locations from (
select adverts.id AS id, service_locations.location_id AS location_id from adverts
left join subscriptions
on adverts.subscription_id = subscriptions.id
left join service_locations
on service_locations.service_id = subscriptions.service_id
order by case service_locations.location_id
when 2 then 1
when 1 then 3
else 2
end
) as table
group by table.id
order by case table.location_id
when 2 then 1
when 1 then 3
else 2
end
Say I have a table of people...
person:
-----------
id | person
---+-------
1 | Jim
2 | Bob
3 | Frank
...and I have a table of items...
item:
----------------
id | item | type
---+------+-----
1 | 21 | 2
2 | 10 | 5
3 | 11 | 1
4 | 9 | 1
...and I also have a table describing who has what...
person_item:
-------------
item | person
-----+-------
1 | 2
2 | 1
3 | 1
How can I create a single query that will tell me when an individual has more than one item of a particular type? I only want the query to concern itself with items of type (1, 2, 3).
The results from the query should be in the following format:
---------------
person | item
| item
--------+------
person | item
| item
| item
--------+------
... etc.
This is what I have tried... but it produces garbage...
SELECT person.id, item.id FROM person_item AS pi
JOIN item AS i ON i.id = pi.item
JOIN person AS p ON p.id = pi.item
WHERE item.type IN (1,2,3)
HAVING COUNT(pi.person) > 1;
The query is suspect because you have a having clause but not a group by clause. Also, you are using table names when you have very reasonable aliases. And, you want to count distinct items within a person/type combination, not just for a person.
Taking these into account, try this query:
SELECT p.id, i.type, group_concat(i.item) as items
FROM person_item pi join
item i
ON i.id = pi.item join
person p
ON p.id = pi.person
WHERE i.type IN (1,2,3)
group by p.id, i.type
HAVING COUNT(distinct i.id) > 1;
This also provides the list of items as the third things returned.
If you only want to see person and item id's, you don't need to join to person - just access person_item (with a link to item for item_type). However, if you want each combination on a separate line, you will have to access person_item twice - like so:
select pi.person, pi.item
from person_item pi
join (select p.person
from person_item p
join item i on p.item = i.item_id and i.type in (1,2,3)
group by p.person
having count(*) > 1) c
on pi.person = c.person
I have two tables, one of items, and one of users who have flagged the items. Here is an example:
items: flags:
item_id | item_name | owner_id item_id | flagged_by
------------------------------ --------------------
1 | foo | 1 1 | 2
2 | bar | 2 2 | 4
3 | baz | 2 2 | 7
2 | 7
I want to select the information from the item table about all the items that are in the flag table, ordered by the number of flags. So for the above example, my desired output would be
item_id | item_name | owner_id
------------------------------
2 | bar | 2
1 | foo | 1
The query I have right now is select * from items where id in (select item_id from flags group by item_id order by count(*) desc);
I know that the inner query works correctly (returns all the IDs in the correct order) but when I run the overall query, I just get the items in order of item id. How do I fix my query?
You are ordering the subquery only currently, which doesn't have an effect on the order of the outer query. If you join the tables rather than using a subquery, you should be able to apply an order to the whole query:
select i.*
from items i
join flags f on i.item_id = f.item_id
group by i.item_id
order by count(f.item_id) desc
Demo: http://www.sqlfiddle.com/#!2/f141b/2