A table (ex: A) has three columns id, name, amount. Amount has some +ve,-ve and zero values. How to write a query to get the amount in such a way that it gives as a result first +ve then -ve then zeros? PFA for the sample table.
Thanks in advance
If you want 3 groups one after the other:
SELECT *
FROM tablename
ORDER BY
(CASE
WHEN amount > 0 then 1
WHEN amount < 0 then 2
ELSE 3
END),
Id;
Are you just looking for order by with a case expression?
select a.*
from a
order by (case when a.amount > 0 then 1
when a.amount < 0 then 2
else 3
end),
a.amount desc;
Related
I have below tables
Table1: "Demo"
Columns: SSN, sales, Create_DT,Update_Dt
Table2: "Agent"
Columns: SSN,sales, Agent_Name, Create_Dt, Update_DT
Scenario 1 and desired result set:
I want output as 0 if the count of SSN in Demo table is matched with the count of SSN in Agent table
if the count is not matched then I want result as 1
Scenario 2 and desired result set:
I want output as 0 if the sum of sales in Demo table is matched with the sum of sales in Agent table
if the sum is not matched then I want result as 1
Please help on this query part
Thanks
You can write two queries separately to take counts within the result query
SELECT (SELECT count(Demo.SSN) as SSN1 from Demo)!=(SELECT count(Agent.SSN) as SSN2 from Agent) AS Result;
Basically what the inner queries does is it checked whether the counts are equal or not and outputs 1 if it is true and 0 if it is false. Since you have asked to output 1 if it is false I used '!=' sign.
You can try the same procedure in scenario 2 also
For scenario 1
select (Case when (select count(ssn) from Demo)=(select count(ssn) from Agent) then 0 else 1 end) as desired_result
If you want to count unique ssn then:
select (Case when (select count(distinct ssn) from Demo)=(select count(distinct ssn) from Agent) then 0 else 1 end) as desired_result
For scenario 2:
select (Case when (select sum(sales) from Demo)=(select sum(sales) from Agent) then 0 else 1 end) as desired_result
I would suggest one query with both sets of information:
select (d.num_ssn <> a.num_ssn) as have_different_ssn_count,
(d.sales <> a.sales) as have_different_sales
from (select count(distinct ssn) as num_ssn,
coalesce(sum(sales), 0) as sales
from demo
) d cross join
(select count(distinct ssn) as num_ssn,
coalesce(sum(sales), 0) as sales
from agent
) a;
Note: This returns boolean values -- true/false rather than 1/0. If you really want 0/1, then use case:
select (case when d.num_ssn <> a.num_ssn then 1 else 0 end) as have_different_ssn_count,
(case when d.sales <> a.sales then 1 else 0 end) as have_different_sales
It would not surprise me if you were not only interested in the total counts but also that the agent/sales combinations are the same in both tables. If that is the case, please ask a new question with a clear explanation. Sample data and desired results help.
I have table whose column is just the length of a session and I would like to return the number of session that have zero length and the number of sessions that have length greater than zero.
I can do that with two separate commands
select count(session_length) from my_table where session_length=0
select count(session_length) from my_table where session_length>0
But I would like to see the results combined in one table
You can do it with one query using conditional aggregation.
select
count(case when session_length = 0 then 1 end),
count(case when session_length > 0 then 1 end)
from my_table
select 1 as QryNo, count(session_length) as SessLen
from my_table
where session_length=0
union
select 2 as QryNo, count(session_length) as SessLen
from my_table
where session_length>0
or
select
case
when session_length = 0 then 1
else 2
end as QryNo,
count(session_length) as SessLen
from my_table
This may be too simple so apologies if I have misread your query but Can you use
select count(session_length) from my_table where session_length >= 0
Again, Apologies if this is not what you're looking for.
My table structure is this
id last_mod_dt nr is_u is_rog is_ror is_unv
1 x uuid1 1 1 1 0
2 y uuid1 1 0 1 1
3 z uuid2 1 1 1 1
I want the count of rows with:
is_ror=1 or is_rog =1
is_u=1
is_unv=1
All in a single query. Is it possible?
The problem I am facing is that there can be same values for nr as is the case in the table above.
Case statments provide mondo flexibility...
SELECT
sum(case
when is_ror = 1 or is_rog = 1 then 1
else 0
end) FirstCount
,sum(case
when is_u = 1 then 1
else 0
end) SecondCount
,sum(case
when is_unv = 1 then 1
else 0
end) ThirdCount
from MyTable
you can use union to get multiple results e.g.
select count(*) from table with is_ror=1 or is_rog =1
union
select count(*) from table with is_u=1
union
select count(*) from table with is_unv=1
Then the result set will contain three rows each with one of the counts.
Sounds pretty simple if "all in a single query" does not disqualify subselects;
SELECT
(SELECT COUNT(DISTINCT nr) FROM table1 WHERE is_ror=1 OR is_rog=1) cnt_ror_reg,
(SELECT COUNT(DISTINCT nr) FROM table1 WHERE is_u=1) cnt_u,
(SELECT COUNT(DISTINCT nr) FROM table1 WHERE is_unv=1) cnt_unv;
how about something like
SELECT
SUM(IF(is_u > 0 AND is_rog > 0, 1, 0)) AS count_something,
...
from table
group by nr
I think it will do the trick
I am of course not sure what you want exactly, but I believe you can use the logic to produce your desired result.
I have a table that has a column called 'status'. This can be set to 0 or 1
Is it possible for me to count both the 0's and 1's in a single query?
Thanks in advance
James
Yes, just group on the value of status:
SELECT status, COUNT(*)
FROM yourtable
GROUP BY status
That will give you exactly two rows since the value can only be 0 or 1, and the COUNT(*) column will be the number of times each status value appears in the table.
SELECT SUM(CASE WHEN status = 0 THEN 1 ELSE 0 END) AS 'number of zeroes',
SUM(CASE WHEN status = 1 THEN 1 ELSE 0 END) AS 'number of ones'
FROM yourtable;
I have one table and one column in it. There is 15 rows (integers). I want to count
the positive numbers and negative numbers, and also sum of total numbers in one query.
Can any one help me?
Or...
SELECT
COUNT(CASE WHEN Col > 0 THEN 1 END) AS NumPositives,
COUNT(CASE WHEN Col < 0 THEN 1 END) AS NumNegatives,
SUM(Col) AS Tot
FROM TableName;
Or you could consider using SIGN(Col), which gives 1 for positive numbers and -1 for negative numbers.
I'll give you psudeo code to help you with your homework.
3 aggregates:
SUM
SUM (CASE < 0)
SUM (CASE > 0)
select (select sum(mycolumn) from mytable where mycolumn > 0) as positive_sum,
(select sum(mycolumn) from mytable where mycolumn < 0) as negative_sum,
sum(mycolumn) as total_sum
from mytable
Try this
SELECT SUM(CASE WHEN Col > 0 THEN 1 ELSE 0 END) AS Pos,
SUM(CASE WHEN Col < 0 THEN 1 ELSE 0 END) AS Neg,
SUM(Col) AS Tot
FROM Table