I want to use SQL to sum multiple rows with same id when wk_days > 10
Sample data :
staff_id wk_days
--------------------
A 5
B 27
B 4
C 13
D 5
Output data :
staff_id wk_days
--------------------
A 5
B 31
C 13
D 5
the above output data is I want, I think I can use CTE to do it. How can it write this SQL Query?
No CTE needed, that is a simple group by query:
select staff_id, sum(wk_days) as wk_days
from the_table
group by staff_id
order by staff_id;
Online example: https://rextester.com/MBE21399
Related
I have a table like the following
ID_A
ID_B
Avg_Class_Size
1
2
16
3
4
10
2
3
8
2
4
9
Where ID_A and ID_B represent distinct student ID codes, and AVG_Class_Size represents the average class size of the classes shared between students A and B.
I would like to calculate the average of the "avg_class_size" for each student, regardless of whether they are student "A" or student "B", with results like below:
ID
AVG
1
16
2
11
3
9
4
9.5
Is there a simple way to accomplish this with a SQL query?
Select with UNION ALL all the ids and averages of the students and aggregate:
SELECT ID, AVG(Avg_Class_Size) average
FROM (
SELECT ID_A ID, Avg_Class_Size FROM tablename
UNION ALL
SELECT ID_B ID, Avg_Class_Size FROM tablename
) t
GROUP BY ID
See the demo.
Results:
ID
average
1
16
2
11
3
9
4
9.5
i have date like this Data
id name period difference
6172 A 6 10
6172 A 3 10
10099 AB 12 24
10099 AB 6 24
10099 AB 3 24
10052 ABC 12 26
10052 ABC 6 26
10052 ABC 3 26
9014 ABCD 12 21
9014 ABCD 6 21
9014 ABCD 3 21
how to get result like this
id name period difference
6172 A 6 10
10099 AB 12 24
10052 ABC 12 26
9014 ABCD 12 4
i try with distinct on (id), but the result like this
id name period difference
6172 A 6 10
10099 AB 6 24
10052 ABC 6 26
9014 ABCD 6 4
The query you want looks something like:
SELECT DISTINCT ON (id) *
FROM Data
ORDER BY id, period DESC;
Demo
This is probably the most efficient way to write your query on Postgres. Note that DISTINCT ON syntax does not support more than one column in the ON clause. The above logic happens to work here assuming that id would uniquely identify each group (that is, that id would always be unique). If not, then we might have to resort to using ROW_NUMBER with a partition over id and name.
using max()
select id, name, t2.period, difference from tableA t1
inner join
(select id, max(period) as period from tableA
group by id) t2 on t2.id = t1.id
using distinct()
select distinct id, name, t2.period, difference from tableA
it seems you need just max()
select id,name,max(period),max(difference)
from table group by id,name
Though i have not found difference=4 in your sample data but you used that on output,so i guessed its your typo
Use max()
select id, name, max(period), difference from tablename
group by id, name,difference
You can try my code:
SELECT
id, name, max(period), difference
FROM
data_table
group by id, name,difference
order by name
This is a demo link http://sqlfiddle.com/#!17/9ab8d/2
I create frequencies on one column in SQL in a standard way.
My code is
select id , count(*) as counts
from TABLE
group by id
order by counts desc
Suppose the output is as follows for six id
id counts
-- -----
1 3 two id have 3 counts per
2 3
---------
3 6 three id have 6 counts per
4 6
5 6
---------
6 2 one id has 2 counts
How can I produce the following?
nid counts
--- ------
1 2
2 3
3 6
I am writing in a hive environment, but that should be standard SQL.
Thanks in advance for answering.
You want two levels of aggregation:
select counts, count(*)
from (select id , count(*) as counts
from TABLE
group by id
) c
group by counts
order by counts;
I call this a "histogram-of-histograms" query. I usually include min(id) and max(id) in the outer select, so I have examples of ids with given frequencies.
I have a table that looks like this:
name | id
-----------
A 1
A 1
B 2
C 1
D 3
D 3
F 2
I want to return id's 1 and 2 because they are duplicate on names. I don't want to return 3, because it is distinct for D 3.
Basically, I'm thinking of doing a query to first get a distinct pairing, so the above reduces to
name | id
-----------
A 1
B 2
C 1
D 3
F 2
And then doing a duplicate find on the id column. However, I'm struggling to find the correct syntax to construct that query.
You should be able to get the result you want by using a GROUP BY along with a HAVING clause that counts the distinct names. The HAVING clause will filter for those ids that have more than one distinct name:
select id
from Table1
group by id
having count(distinct name) > 1
Here is a demo
I have a table like in example below.
SQL> select * from test;
ID PARENT_ID NAME
1 1 A
2 1 B
3 2 A
4 2 B
5 3 A
6 3 B
7 3 C
8 4 A
What I need is to get all unique subsets of names ((A,B), (A,B,C), (A)) or exclude duplicate subsets. You can see that (A,B) is twice there, one for PARENT_ID=1 and one for 2.
I want to exclude such duplicates:
ID PARENT_ID NAME
1 1 A
2 1 B
5 3 A
6 3 B
7 3 C
8 4 A
You can use DISTINCT to only return different values.
e.g.
SELECT DISTINCT GROUP_CONCAT(NAME SEPARATOR ',') as subsets
FROM TABLE_1
GROUP BY PARENT_ID;
SQL Fiddle
I have used 'group_concat' assuming you are using 'Mysql'. The equivalent function in Oracle is 'listagg()'. you can see it in action here in SQL fiddle
Here is the solution:-
Select a.* from
test a
inner join
(
Select nm, min(parent_id) as p_id
from
(
Select Parent_id, group_concat(NAME) as nm
from test
group by Parent_ID
) a
group by nm
)b
on a.Parent_id=b.p_id
order by parent_id, name