I have a table of data, see
Using a pivot statement, I am able to break down the count by title
select * from (
select * from ta
)
pivot (
COUNT(title)
for title in ( 'worker', 'manager') )
So the result looks like this:
STATUS 'worker' 'manager'
started 3 1
finished 4 5
ready 3 4
What I need to add a third column for the row totals
STATUS 'worker' 'manager' Total
started 3 1 4
finished 4 5 9
ready 3 4 7
Any idea how I can accomplish this within the same statement?
demo is at http://sqlfiddle.com/#!4/740fd/1
I would just use conditional aggregation rather than pivot. This gives you the extra flexibility that you need:
select
status,
sum(case when title = 'worker' then 1 else 0 end) worker,
sum(case when title = 'manager' then 1 else 0 end) manager,
count(*) total
from ta
group by status
Demo on DB Fiddle:
STATUS | WORKER | MANAGER | TOTAL
:------- | -----: | ------: | ----:
started | 3 | 1 | 4
finished | 4 | 5 | 9
ready | 3 | 4 | 7
Use the SUM() analytic function to get the total and then use PIVOT
select
status,
sum(case
when title = 'worker'
then 1
else 0
end) worker,
sum(case
when title = 'manager'
then 1
else 0
end) manager,
count(*) total
from ta
group by status
Give an alias for the whole query(such as q) in order to qualify the all columns with asterisk(q.*), and then sum up all the columns to yield total column next to it :
select q.*, worker + manager as total
from ta
pivot
(
count(title)
for title in ( 'worker' as worker, 'manager' as manager )
) q
Demo
I think the other examples are much simpler, but here is a different approach using cube and grouping before pivoting:
select *
from (
select decode(grouping(title),1,'total',0,title) title,
status,
count(*) cnt
from ta
group by status, cube(title) )
pivot(
sum(cnt) for title in ('worker','manager','total')
)
Output:
| STATUS | 'worker' | 'manager' | 'total' |
|----------|----------|-----------|---------|
| finished | 4 | 5 | 9 |
| ready | 3 | 4 | 7 |
| started | 3 | 1 | 4 |
http://sqlfiddle.com/#!4/740fd/13/0
Adding the cube into the group by clause will give you a subtotal for that column. It will show as null in that column by default. You can use the grouping function in the select clause to differentiate between the total row and the normal rows (the total row will be 1, normal rows are 0). Using a decode will force those total rows to be 'total' which becomes one of the values that you can pivot on.
Related
I have a problem with grouping records in PostgreSQL. I have a structure containing 3 columns, non unique id, name, group (it's old system and I can't change this structure).
Sample records:
id | name | group
-----+----------+------
1 | product1 | 0
1 | product1 | test
2 | product2 | test
3 | product3 | test123
I want the groups unequal 0 to be concatenated (get the id, name of the first record from the group).
The expected result:
id | name | group
-----+----------+------
1 | product1 | 0
1 | product1 | test
3 | product3 | test123
Currently count records in the following way:
SELECT
COUNT(CASE WHEN group = '0' THEN group END) +
COUNT(DISTINCT CASE WHEN group <> '0' THEN group END) AS count
FROM
table
Is it correct way? How can I convert it to retrieve records?
You can use row_number():
select id, name, group
from (select t.*, row_number() over (partition by group order by id) as seqnum
from t
) t
where seqnum = 1 or group = '0';
Note: group is a really bad name for a column. It is a SQL keyword, so you should escape the name. I am leaving it as is because your query uses it.
I'm trying to formulate a query to aggregate rows that are between rows with a specific value: in this example I want to collapse and sum time of all rows that have an ID other than 1, but still show rows with ID 1.
This is my table:
ID | Time
----+-----------
1 | 60
2 | 10
3 | 15
1 | 30
4 | 100
1 | 20
This is the result I'm looking for:
ID | Time
--------+-----------
1 | 60
Other | 25
1 | 30
Other | 100
1 | 20
I have attempted to SUM and add a condition with CASE, or but so far my solutions only get me to sum ALL rows and I lose the intervals, so I get this:
ID | Time
------------+-----------
Other | 125
1 | 110
Any help or suggestions in the right direction would be greatly appreciated, thanks!
You need to define the groupings. SQLite is not great for this sort of manipulation, but you can do it by summing the "1" values up to each value.
In SQLite, we can use the rowid column for the ordering:
select (case when id = 1 then '1' else 'other' end) as which,
sum(time)
from (select t.*,
(select count(*) from t t2 where t2.rowid <= t.rowid and t2.id = 1) as grp
from t
) t
group by (case when id = 1 then '1' else 'other' end), grp
order by grp, which;
I have created a table from multiple tables like this:
Week | Cid | CustId | L1
10 | 1 | 1 | 2
10 | 2 | 1 | 2
10 | 5 | 1 | 2
10 | 4 | 1 | 1
10 | 3 | 2 | 1
4 | 6 | 1 | 2
4 | 7 | 1 | 2
I want the output as:
Repeat
0
1
1
0
0
0
1
So, basically what I want is for each week, if a person (custid) comes in again with the same L1, then the value in the column Repeat should become 1, otherwise 0 ( so like, here, in row 2 & 3, custid 1, came with L1=2 again, so it will get 1 in column "Repeat", however in row 4, custid 1 came with L1=1, so it will get value as ).
By the way, the table isn't ordered (as I've shown).
I'm trying to do it as follows:
select t.*,
lag(0, 1, 0) over (partition by week, custid, L1 order by cid) as repeat
from
table;
But this is not giving the output and is giving empty result.
I think you need a case, but I would use row_number() for this:
select t.*,
(case when row_number() over (partition by week, custid, l1 order by cid) = 1
then 0 else 1
end) as repeat
from table;
This can also be computed without Window functions but by a self-join in the following way:
SELECT a.week, a.cid, a.custid, a.l1,
CASE WHEN b IS NULL THEN 1 ELSE 0 END AS repeat
FROM mytable a NATURAL LEFT JOIN
(SELECT week, min(cid) AS cid, custid, l1 FROM mytable
GROUP BY week,custid,l1) b
ORDER BY week DESC, custid, l1 DESC, cid;
It can be done simply by using an count(*) as analytic function. No case expression or self join needed. The query is even portable across databases that support analytic functions:
SELECT cust.*, least(count(*)
OVER (PARTITION BY Week, CustId, L1 ORDER BY Cid
ROWS UNBOUNDED PRECEDING) - 1, 1) repeat
FROM cust ORDER BY Week DESC, custId, L1 DESC;
Executing the query on your data results in the following output (last row is the repeat row):
Week | Cid | CustId | L1 | repeat
10 1 1 2 0
10 2 1 2 1
10 5 1 2 1
10 4 1 1 0
10 3 2 1 0
4 6 1 2 0
4 7 1 2 1
Tested on Oracle 11g and PostgreSQL 9.4. Note that the second ORDER BY is optional. See Oracle Language Reference, Analytic Functions for more details.
I want to create an additional column which calculates the value of a row from count column with its predecessor row from the sum column. Below is the query. I tried using ROLLUP but it does not serve the purpose.
select to_char(register_date,'YYYY-MM') as "registered_in_month"
,count(*) as Total_count
from CMSS.USERS_PROFILE a
where a.pcms_db != '*'
group by (to_char(register_date,'YYYY-MM'))
order by to_char(register_date,'YYYY-MM')
This is what i get
registered_in_month TOTAL_COUNT
-------------------------------------
2005-01 1
2005-02 3
2005-04 8
2005-06 4
But what I would like to display is below, including the months which have count as 0
registered_in_month TOTAL_COUNT SUM
------------------------------------------
2005-01 1 1
2005-02 3 4
2005-03 0 4
2005-04 8 12
2005-05 0 12
2005-06 4 16
To include missing months in your result, first you need to have complete list of months. To do that you should find the earliest and latest month and then use heirarchial
query to generate the complete list.
SQL Fiddle
with x(min_date, max_date) as (
select min(trunc(register_date,'month')),
max(trunc(register_date,'month'))
from users_profile
)
select add_months(min_date,level-1)
from x
connect by add_months(min_date,level-1) <= max_date;
Once you have all the months, you can outer join it to your table. To get the cumulative sum, simply add up the count using SUM as analytical function.
with x(min_date, max_date) as (
select min(trunc(register_date,'month')),
max(trunc(register_date,'month'))
from users_profile
),
y(all_months) as (
select add_months(min_date,level-1)
from x
connect by add_months(min_date,level-1) <= max_date
)
select to_char(a.all_months,'yyyy-mm') registered_in_month,
count(b.register_date) total_count,
sum(count(b.register_date)) over (order by a.all_months) "sum"
from y a left outer join users_profile b
on a.all_months = trunc(b.register_date,'month')
group by a.all_months
order by a.all_months;
Output:
| REGISTERED_IN_MONTH | TOTAL_COUNT | SUM |
|---------------------|-------------|-----|
| 2005-01 | 1 | 1 |
| 2005-02 | 3 | 4 |
| 2005-03 | 0 | 4 |
| 2005-04 | 8 | 12 |
| 2005-05 | 0 | 12 |
| 2005-06 | 4 | 16 |
I have a Table with different Cluster ID's
ID
1
1
2
2
2
3
3
3
3
4
4
I want to display the size of the Cluster with No of Clusters in that Cluster.
For example for the above table Expected Output:
Cluster Size | No of Clusters (with that size)
2 | 2
3 | 1
4 | 1
I wrote a query which will give me the specified Cluster size.
Select COUNT(*) from
(SELECT ID, COUNT(ID) as cnt
FROM [Table] group by ID having COUNT(*) =3) as TC;
In the above example I will get "1" as my result for the above table.
However, I want a query which will give me all the Clusters and their respective size.
select [Cluster Size], Count(*) as [No of Clusters]
from (
select count(*) as [Cluster Size]
from Table1
group by ID
) a
group by [Cluster Size]
SQL Fiddle Example
Output:
| CLUSTER SIZE | NO OF CLUSTERS |
---------------------------------
| 2 | 2 |
| 3 | 1 |
| 4 | 1 |
SELECT c AS ClusterSize, COUNT(*) AS NumOfClusters
FROM
(
SELECT COUNT(*) AS c, ID
FROM #table
GROUP BY ID
)A
GROUP BY c
If I understood your requirements this query should do the job:
SELECT cl.size, COUNT(cl.size)
FROM
(SELECT id, COUNT(id) [Size] FROM Table1 GROUP BY id) cl
GROUP BY cl.size
Here is a link to SQL Fiddle: http://sqlfiddle.com/#!3/56de3/7