I want to make a select query which groups rows based on a given column and then sorts by size of such groups.
Let's say we have this sample data:
id type
1 c
2 b
3 b
4 a
5 c
6 b
I want to obtain the following by grouping and sorting the column 'type' in a descending way:
id type
2 b
3 b
6 b
1 c
5 c
4 a
As of now I am only able to get the count of each group but that is not exactly what I need:
SELECT *, COUNT(type) AS typecount
FROM sampletable
GROUP BY type
ORDER BY typecount DESC, type ASC
id type count
2 b 3
1 c 2
4 a 1
Can anybody please give me a hand with this query?
Edit:
Made 'b' the biggest group to avoid coming to the same solution by using only SORT BY
You can't use a column alias in your GROUP BY; just repeat the expression:
SELECT type, COUNT(type) AS count
FROM sampletable
GROUP BY type
ORDER BY COUNT(*) DESC, type ASC
Note that I changed the SELECT clause - you can't use * in your SELECT either since expressions in the SELECT need to either be in the GROUP BY clause or an aggregation.
It may not be the best way, but it will give you what you want.
You work out the totals for each group and then join that "virtual" table to your original table by the determined counts.
SELECT *
FROM sampletable s1
INNER JOIN (SELECT count(type) AS iCount,type
FROM sampletable
GROUP BY type) s2 ON s2.type = s1.type
ORDER BY s2.iCount DESC, s1.type ASC
http://sqlfiddle.com/#!9/f6b0c4/6/0
You can't perform GROUP BY operation on COLUMN ALIAS.
The reason why you can't use ALIAS on the GROUP BY clause that is created on the same level of the SELECT statement is because the GROUP BY is executed before the SELECT clause in which the ALIAS is created.
This is the SQL Order of Operation:
FROM clause
WHERE clause
GROUP BY clause
HAVING clause
SELECT clause
ORDER BY clause
Try following query:
SELECT type, COUNT(type) AS count
FROM sampletable
GROUP BY type
ORDER BY COUNT(*) DESC, type ASC;
EDIT:-
SELECT id, type
FROM sampletable
ORDER BY type DESC, id ASC;
Related
When using Array_agg, it returns the same values in different orders. I tried using distinct in a few places and it didn't work. I tried using an order before and after the array and it would fail or not properly exclude results.
I am trying to find all fields in the field column that share the same time and same ID and put them into an array.
Columns are Fieldname, ID, Time
select b.Field, count(*)
from (select Time, ID, array_agg(fieldname) as Field
from a
group by 1,2
order by 3) b
group by b.field
order by 1 desc
This produces duplicate results
For example I will have:
Field Name Count
Ghost,Mark 1234
Mark,Ghost 1234
I also tried this below where I add a subquery where I first order the fields alphabetically when grouping time and ID but it failed to execute. I think due to array_agg not being the root query?
select a.Field, count(*)
from
(select Time, ID, array_agg(fieldname) as field
from
(select Time, ID, fieldname
from a
group by 1,2
order by 3 desc) a
group by 1,2 ) b
group by 1
order by 2 desc
I have a table in postgres with 2 fields: they are columns of ids of users who have looked at some data, under two conditions:
viewee viewer
------ ------
93024 66994
93156 93151
93163 113671
137340 93161
92992 93161
93161 93135
93156 93024
And I want to group them by both viewee and viewer field, and count the number of occurrences, and return that count
from high to low:
id count
------ -----
93161 3
93156 2
93024 2
137340 1
66994 1
92992 1
93135 1
93151 1
93163 1
I have been running two queries, one for each column, and then combining the results in my JavaScript application code. My query for one field is...
SELECT "viewer",
COUNT("viewer")
FROM "public"."friend_currentfriend"
GROUP BY "viewer"
ORDER BY count DESC;
How would I rewrite this query to handle both fields at once?
You can combine to columns from the table into a single one by using union all then use group by as below:
select id ,count(*) Count from (
select viewee id from vv
union all
select viewer id from vv) t
group by id
order by count(*) desc
Results:
This is a good place to use a lateral join:
select v.viewx, count(*)
from t cross join lateral
(values (t.viewee), (t.viewer)) v(viewx)
group by v.viewx
order by count(*) desc;
You can try this :
SELECT a.ID,
SUM(a.Total) as Total
FROM (SELECT t.Viewee AS ID,
COUNT(t.Viewee) AS Total
FROM #Temp t
GROUP BY t.Viewee
UNION
SELECT t.Viewer AS ID,
COUNT(t.Viewer) AS Total
FROM #Temp t
GROUP BY t.Viewer
) a
GROUP BY a.ID
ORDER BY SUM(a.Total) DESC
I have this table called item:
| PERSON_id | ITEM_id |
|------------------|----------------|
|------CP2---------|-----A03--------|
|------CP2---------|-----A02--------|
|------HB3---------|-----A02--------|
|------BW4---------|-----A01--------|
I need an SQL statement that would output the person with the most Items. Not really sure where to start either.
I advice you to use inner query for this purpose. the inner query is going to include group by and order by statement. and outer query will select the first statement which has the most items.
SELECT * FROM
(
SELECT PERSON_ID, COUNT(*) FROM TABLE1
GROUP BY PERSON_ID
ORDER BY 2 DESC
)
WHERE ROWNUM = 1
here is the fiddler link : http://sqlfiddle.com/#!4/4c4228/5
Locating the maximum of an aggregated column requires more than a single calculation, so here you can use a "common table expression" (cte) to hold the result and then re-use that result in a where clause:
with cte as (
select
person_id
, count(item_id) count_items
from mytable
group by
person_id
)
select
*
from cte
where count_items = (select max(count_items) from cte)
Note, if more than one person shares the same maximum count; more than one row will be returned bu this query.
Hope someone can help.
I have been trying a few queries but I do not seem to be getting the desired result.
I need to identify the highest ‘’claimed’’ users within my table without discarding the columns from the final report.
The user can have more than one record in the table, however the data will be completely different as only the user will match.
The below query only provides me the count per user without giving me the details.
SELECT User, count (*) total_record
FROM mytable
GROUP BY User
ORDER BY count(*) desc
Table:
mytable
Column 1 = User Column 2 = Ref Number Column 3 = Date
The first column will be the unique identifier, however the data in the other columns will differ, therefore it needs to descend the highest claimed user with all the relevant rows to the user to the least claimed user.
User|Ref Num|Date
1|a|20150317
1|b|20150317
2|c|20150317
3|d|20150317
4|e|20150317
1|f|20150317
4|e|20150317
The below data is how the values should be returned.
User|Ref Num|Date|Count
1|a|20150317|3
1|b|20150317|3
1|f|20150317|3
2|c|20150317|1
3|d|20150317|1
4|e|20150317|2
4|e|20150317|2
Hope it makes sense.
Thank you
As you're using MSSQL you can use the OVER() clause like so:
SELECT [user], mt.ref_num, mt.[date], COUNT(mt.[user]) OVER(PARTITION BY mt.[user])
FROM myTable mt
More about the OVER clause can be found here: https://msdn.microsoft.com/en-us/library/ms189461.aspx
As per your comment you can use the wildcard * like so:
SELECT mt.*, COUNT(mt.[user]) OVER(PARTITION BY mt.[user])
FROM myTable mt
This would get you every column as well as the result of the count.
If you want to order by the number of record for each user, then use window functions instead of aggregation:
SELECT t.*
FROM (SELECT t., count(*) OVER (partition by user) as cnt
FROM mytable t
) t
ORDER BY cnt DESC, user;
Note that I added user to the order by so users with the same count will appear together in the list.
You could use an outer apply if your version of SQL Server supports it:
SELECT [User], [Ref Num], Date, total_record
FROM mytable M
OUTER APPLY (
SELECT count(*) total_record
FROM mytable
WHERE [user] = M.[user]
GROUP BY [user]
) oa
ORDER BY total_record desc, [user]
Note that user is a reserved keyword in MSSQL and you need to enclose it in either brackets [user] or double-quotes "user".
This would produce an output like:
user Ref Num Date total_record
1 a 2015-03-17 3
1 b 2015-03-17 3
1 f 2015-03-17 3
4 e 2015-03-17 2
4 e 2015-03-17 2
2 c 2015-03-17 1
3 d 2015-03-17 1
Note that the answers using the count(*) OVER (partition by [user]) construct are more efficient though.
Most simple way would be to use window fuction.
SELECT table.*, COUNT(*) OVER (PARTITION BY user)
FROM nameoftable table -- this is an alias
ORDER BY user, ref_num
This also seem to fit your need.
This is the old way of doing it. Where possible you should use OVER but as other people have answered with that I thought I'd throw this one into the mix.
SELECT
T.[User]
,T.[Ref Num]
,T.[Date]
,(SELECT count(*) from [myTable] T2 where T2.[User] = T.[USER]) as [Count]
FROM [mytable] T
ORDER BY [Count] DESC
I am trying to find a query that would give me a count of another table in the query. The problem is that I have no idea what to set where in the count part to. As it is now it will just give back a count of all the values in that table.
Select
ID as Num,
(select Count(*) from TASK where ID=ID(Also tried Num)) as Total
from ORDER
The goal is to have a result that reads like
Num Total
_________________
1 13
2 5
3 22
You need table aliases. So I think you want:
Select ID as Num,
(select Count(*) from TASK t where t.ID = o.ID) as Total
from ORDER o;
By the way, ORDER is a terrible name for a table because it is a reserved work in SQL.
You can do it as a sub query or a join (or an OVER statement.)
I think the join is clearest when you are first learning SQL
Select
ID as Num, count(TASK.ID) AS Total
from ORDER
left join TASK ON ORDER.ID=TASK.ID
GROUP BY ORDER.ID