This question already has answers here:
How to find the maximum count using mysql?
(9 answers)
Closed 1 year ago.
I've been trying to get the maximum number from a set of different values.
suppose,
i have table with values,
id val
A 1
A 2
A 3
B 4
B 5
C 6
so i want to get the maximum number of id and first i did get the count and tried to get maximum
my sql query is,
SELECT MAX (ID_COUNT) id_count
FROM (SELECT id,COUNT(id) ID_COUNT
FROM tab1
GROUP BY id)
I want an ouput 3 since A is repeated 3 times and is the maximum.
it was supposed to work. but now im getting an error saying derived table value should have an alias. but i did give an alias.
what should i do?
thanks in advance...
You need to specify an alias for the sub select
SELECT MAX (ID_COUNT) id_count
FROM (SELECT id,COUNT(id) ID_COUNT
FROM tab1
GROUP BY id) tab2
You can try the below -
SELECT id,COUNT(id) ID_COUNT
FROM tab1
GROUP BY id
order by ID_COUNT desc
limit 1
Related
This question already has answers here:
SQL select only rows with max value on a column [duplicate]
(27 answers)
How can I SELECT rows with MAX(Column value), PARTITION by another column in MYSQL?
(22 answers)
MySQL - How to select rows with max value of a field
(3 answers)
Closed 1 year ago.
I'm looking for a way to display a column in my table without aggregating it (column store)
SELECT id, name, max(cost), store
FROM test
group by 1,2,4
But in the table im using there are some row that shares the same id,name,license with different machine, and Ideally I'd like to get in return only a single row, with max(cost), if they share the same first 2 columns and only display the last column - without using it to group by.
actual outcome:
id
name
max(cost)
store
1
Joe
30
store1
1
Joe
50
store2
but my desired result will be:
id
name
max(cost)
store
1
Joe
50
store2
Can it even be done?
You seem to want the row with the maximum cost. A typical method uses row_number();
select t.*
from (select t.*,
row_number() over (partition by id, name order by cost desc) as seqnum
from test t
) t
where seqnum = 1;
No aggregation is needed.
You can also use a correlated subquery:
select t.*
from test t
where t.cost = (select max(t2.cost)
from test t2
where t2.id = t.id
);
This question already has answers here:
SQL select only rows with max value on a column [duplicate]
(27 answers)
Closed 2 years ago.
I have the following query.
SELECT MAX(activity_id), group_id, parent_group_id FROM dashboard_activity_detail WHERE account_id = 8997 AND parent_group_id IN (5118,5026,4522,3983,3586,3278,3227) AND activity_type_id = 18 GROUP BY parent_group_id;
My expectation is that the group_id will be returned associated with the largest activity_id. There are multiple rows with the same parent_group_id in the table. What I get back is a value from a different row. The activity_id is correct and the parent_group_id is correct but I get a value from a different row for the group_id.
What am I missing? I've tried order by and various other methods with the same result.
You expectation is wrong. You query is malformed because the SELECT columns are inconsistent with the GROUP BY columns.
Use window functions (available in MySQL 8+):
SELECT da.*
FROM (SELECT da.*,
ROW_NUMBER() OVER (PARTITION BY parent_group_id ORDER BY activity_id DESC) as seqnum
FROM dashboard_activity_detail da
WHERE account_id = 8997 AND
parent_group_id IN (5118,5026,4522,3983,3586,3278,3227) AND
activity_type_id = 18
) da
WHERE seqnum = 1;
This question already has answers here:
How to count occurrences of a column value efficiently in SQL?
(7 answers)
Closed 5 years ago.
I have got a table with ID's:
Id
===
1
2
2
3
1
2
And I want to create a table with two columns like this:
Id COUNT
=============
1 2
2 3
3 1
How can I do that?
Let's say you called your table 'user', you can try this :
SELECT user.Id as ID, count(user.Id) as COUNT_ID
FROM user
GROUP BY ID;
Hope it helps,
WaLinke
You have to group by your id.
Select id, count(1) as COUNT
from yourtable
group by ID
order by id
That way you tell your sql, that you want to count the number of rows per id.
If you need more examples feel free to google sql count.
There are many good examples out there.
Or check this stackoverflow question:
How to use count and group by at the same select statement
You can use GROUP BY and Count(columnname) functions like this
SELECT
Id,
Count(Id) AS COUNT
FROM
tablename
GROUP BY Id
Something like this should work
SELECT id, COUNT(id) AS Expr1 FROM dbo.Table1 GROUP BY id
I got a question in my homework for SQL about selecting the maximum values from the same table that have different class "Letters"
For example:
ID Student Group Avg(value)
-------------------------------------
1 stud1 A 9
2 stud2 A 9.5
3 stud3 B 8
4 stud4 B 8.5
What my query should do, is to show stud2 and stud4.The maximum from their respective groups.
I managed to do it in the end, but it took a lot of characters so I thought that maybe there's a shorter way to do. Any ideas? I used to first search the id or the stud that has max avg(value) from group A, intersecting with the id of the stud that has max avg(value) from B and then putting everything into one big select and then using those intersected IDs into another query that requested to show some different things about those IDs. But as I said, it looked far too long and thought that maybe there's an shorter way.
Try this (I renamed group to grp and avg to avg_val as those are reserved keywords):
select t1.*
from your_table t1
inner join (
select grp, max(avg_val) avg_val
from your_table
group by grp
) t2 on t1.grp = t2.grp
and t1.avg_val = t2.avg_val;
It finds maximum avg value per group and joins it with original table to get the corresponding students.
Please note that if there are multiple students with same avg as the max value of the that group, all of those students will be returned.
I have table that looks like this:
id rank
a 2
a 1
b 4
b 3
c 7
d 1
d 1
e 9
I need to get all the distinct rank values on one column and count of all the unique id's that have reached equal or higher rank than in the first column.
So the result I need would be something like this:
rank count
1 5
2 4
3 3
4 3
7 2
9 1
I've been able to make a table with all the unique id's with their max rank:
SELECT
MAX(rank) AS 'TopRank',
id
FROM myTable
GROUP BY id
I'm also able to get all the distinct rank values and count how many id's have reached exactly that rank:
SELECT
DISTINCT TopRank AS 'rank',
COUNT(id) AS 'count of id'
FROM
(SELECT
MAX(rank) AS 'TopRank',
id
FROM myTable
GROUP BY id) tableDerp
GROUP BY TopRank
ORDER BY TopRank ASC
But I don't know how to get count of id's where the rank is equal OR HIGHER than the rank in column 1. Trying SUM(CASE WHEN TopRank > TopRank THEN 1 END) naturally gives me nothing. So how can I get the count of id's where the TopRank is higher or equal to each distinct rank value? Or am I looking in the wrong way and should try something like running totals instead? I tried to look for similar questions but I think I'm completely on a wrong trail here since I couldn't find any and this seems a pretty simple problem that I'm just overthinking somehow. Any help much appreciated.
One approach is to use a correlated subquery. Just get the list of ranks and then use a correlated subquery to get the count you are looking for:
SELECT r.rank,
(SELECT COUNT(DISTINCT t2.id)
FROM myTable t2
WHERE t2.rank >= r.rank
) as cnt
FROM (SELECT DISTINCT rank FROM myTable) r;