I have a table what counts admited and discharge from ED. This is what the table looks like
I'm trying to get the total admitted and discharged. Something like
Admitted - 100
Discharge - 200
Is there a way to do that with the NULL values?
select
sum(case when Admitted is null then 0 else Admitted end) Admitted,
sum(case when DischargedFromED is null then 0 else DischargedFromED end) as DischargedFromED
from MyMagicalTable;
or
select
sum(coalesce(Admitted, 0)) Admitted,
sum(coalesce(DischargedFromED, 0)) DischargedFromED
from MyMagicalTable;
Just use sum();
select sum(admitted) as admitted, sum(DischargedFromED) as DischargedFromED
from t;
Aggregation functions ignore NULL values.
If you are concerned about NULL values appearing after the sum(), then use coalesce() afterwards:
select coalesce(sum(admitted), 0) as admitted,
coalesce(sum(DischargedFromED), 0) as DischargedFromED
from t;
The above assumes that the columns are numeric. If they are some other type, they need to be converted to numbers.
Related
I have a multiple ctes. In my select statement I must filter values base on the conditions. This is my query.
SELECT roadName
,sum(roadLength) AS sumRoadLength
,avg(elevationDifference) AS eglAvgDepth
,avg(elevationDifference) AS pglAvgDepth
,
FROM cte3
GROUP BY roadName
ORDER BY roadName
Under "elevationDifference" there are lots of values ranging from -10 to +20 which are spread through "roadName". What i want to accomplished is that "eglAvgDepth" will return if all "elevationDifference" values are <0 and take the average. Same case with pglAvgDepth where values are >0.
I tried to add where statement but works only in eglAvgDepth
WHERE elevationDifference < 0
GROUP BY roadName
ORDER BY roadName
Just add a conditional expression:
avg(case when elevationDifference < 0 then elevationDifference end) as eglAvgDepth,
avg(case when elevationDifference > 0 then elevationDifference end) as pglAvgDepth,
EDIT:
You have phrased this that you want the value based on whether all the values are positive or negative. If so:
(case when max(elevationDifference) < 0 then avg(elevationDifference) end) as eglAvgDepth,
(case when max(elevationDifference) > 0 then avg(elevationDifference) end) as pglAvgDepth,
I searched for many solutions on SO and elsewhere but couldn't quite understand how to write a query for my problem.
Anyway my query looks like below
SELECT * FROM
(
SELECT Id, Date, Name, Amount,
CASE
WHEN DATEDIFF(DAY,Date,GETDATE()) <=0
THEN 'Current'
WHEN DATEDIFF(DAY,Date,GETDATE()) <30
THEN 'Due30'
WHEN DATEDIFF(DAY,Date,GETDATE()) <60
THEN 'Due60'
ELSE 'Due90'
END AS [Age]
FROM Statement
WHERE (Amount <> 0)
) AS S
PIVOT
(
SUM(Amount)
FOR[Age] IN ([Current],[Due30],[Due60],[Due90])
) P
and the result looks like this
Id Date Name Current Due30 Due60 Due90
----------- ---------- --------------------------------------------
1 2016-04-03 Alan NULL NULL NULL 110.00
2 2016-05-02 TC NULL NULL 30.00 NULL
where should i insert IsNull condition to be able to remove the null in the result and add a zero there.
I tried inserting IsNull in the pivot query but we all know that is not meant to work
You have to add it repetitively in the final SELECT, when you replace the SELECT * (which should only exist in ad-hoc queries or EXISTS tests) with the column list:
SELECT
Id,
Date,
Name,
COALESCE([Current],0) as [Current],
COALESCE(Due30,0) as Due30,
COALESCE(Due60,0) as Due60,
COALESCE(Due90,0) as Due90
FROM
(
SELECT Id, Date, Name, Amount,
CASE
WHEN DATEDIFF(DAY,Date,GETDATE()) <=0
THEN 'Current'
WHEN DATEDIFF(DAY,Date,GETDATE()) <30
THEN 'Due30'
WHEN DATEDIFF(DAY,Date,GETDATE()) <60
THEN 'Due60'
ELSE 'Due90'
END AS [Age]
FROM Statement
WHERE (Amount <> 0)
) AS S
PIVOT
(
SUM(Amount)
FOR[Age] IN ([Current],[Due30],[Due60],[Due90])
) P
I've also used COALESCE since it's generally the preferred option (ANSI standard, extends to more than two arguments, applies normal type precedence rules) instead of ISNULL.
SELECT Id
, [Date]
, Name
, [Current] = SUM(CASE WHEN val <= 0 THEN Amount ELSE 0 END)
, Due30 = SUM(CASE WHEN val < 30 THEN Amount ELSE 0 END)
, Due60 = SUM(CASE WHEN val < 60 THEN Amount ELSE 0 END)
, Due90 = SUM(CASE WHEN val >= 60 THEN Amount ELSE 0 END)
FROM dbo.[Statement] t
CROSS APPLY (
SELECT val = DATEDIFF(DAY, [Date], GETDATE())
) s
WHERE Amount <> 0
GROUP BY Id, [Date], Name
I have some fields in the table, need to average those fields.
Then I run this syntax, because I don't want to calculate 0 (zero) value.
SELECT myDate, AVG(CASE myField1 WHEN 0 THEN NULL ELSE myField1 END) AS avgmyField1
FROM myTable WHERE myDate = '2014-06-01'
On my syntax, the average calculation means.. Make zero value to null.
My question is, How if all values are zero...?
Thank you.
Then you get NULL.
If you want zero instead, use COALESCE:
COALESCE( AVG(CASE myField1 WHEN 0 THEN NULL ELSE myField1 END) , 0)
Average will not use nullvalues to calculate an average value
IsNull or Coalesce can be used to change null values to different values.
This script will change 0 to null and take the average value:
SELECT IsNull(AVG(NullIf(val,0)), 0)
FROM
(Values(5),(7),(0)) tbl(val)
Since 0 is excluded the result is 6
You could also do this:
SELECT myDate, COALESCE(AVG(myField1), 0) AS avgmyField1
FROM myTable WHERE myDate = '2014-06-01' and myField1 <> 0
I would like to do something like this:
select sum(nvl(total_time_out, 0)),
sum(nvl((case when day_of_week = 'Mon' then total_time_out else 0 end) over (partition by person_id), 0))
from xxpay_tna_summary_v
where person_id = 7926
where the second column only returns the sum of the total time out hours for the Monday. Is this possible in Oracle SQL, and what is the correct syntax?
check this
http://sqlfiddle.com/#!4/df376/2
select sum((case when person_id = 100 then total_time_out else 0 end)) total_time,
sum(nvl((case when day_of_week = 'MON' then total_time_out else 0 end), 0)) monday_time
from xxpay_tna_summary_v
Your syntax is invalid, because sum belongs to over, but you moved the sum keyword to the beginning of the expression. Here is the corrected statement:
select nvl(sum(total_time_out), 0),
nvl(sum(case when day_of_week = 'Mon' then total_time_out else 0 end) over (partition by person_id), 0)
from xxpay_tna_summary_v
where person_id = 7926;
(I also changed places for sum and nvl in your first expression. It does the same but might be nanoseconds faster, because nvl has to be applied just once.)
how to calculate count based on rows?
SOURCE TABLE
each employee can take 2 days off
Employee-----First_Day_Off-----Second_Day_Off
1------------10/21/2009--------12/6/2009
2------------09/3/2009--------12/6/2009
3------------09/3/2009--------NULL
4
5
.
.
.
Now i need a table that shows the dates and number of people taking off on that day
Date---------First_Day_Off-------Second_Day_Off
10/21/2009---1-------------------0
12/06/2009---1--------------------1
09/3/2009----2--------------------0
Any ideas?
Oracle 9i+, using Subquery Factoring (WITH):
WITH sample AS (
SELECT a.employee,
a.first_day_off AS day_off,
1 AS day_number
FROM YOUR_TABLE a
WHERE a.first_day_off IS NOT NULL
UNION ALL
SELECT b.employee,
b.second_day_off,
2 AS day_number
FROM YOUR_TABLE b
WHERE b.second_day_off IS NOT NULL)
SELECT s.day_off AS date,
SUM(CASE WHEN s.day_number = 1 THEN 1 ELSE 0 END) AS first_day_off,
SUM(CASE WHEN s.day_number = 2 THEN 1 ELSE 0 END) AS second_day_off
FROM sample s
GROUP BY s.day_off
Non Subquery Version
SELECT s.day_off AS date,
SUM(CASE WHEN s.day_number = 1 THEN 1 ELSE 0 END) AS first_day_off,
SUM(CASE WHEN s.day_number = 2 THEN 1 ELSE 0 END) AS second_day_off
FROM (SELECT a.employee,
a.first_day_off AS day_off,
1 AS day_number
FROM YOUR_TABLE a
WHERE a.first_day_off IS NOT NULL
UNION ALL
SELECT b.employee,
b.second_day_off,
2 AS day_number
FROM YOUR_TABLE b
WHERE b.second_day_off IS NOT NULL) s
GROUP BY s.day_off
It is a bit awkward to handle these queries, since you have days off stored in different columns. A better layout would be to have something like
EMPLOYEE_ID DAY_OFF
Then you would have multiple rows if an employee took multiple days off
EMPLOYEE_ID DAY_OFF
1 10/21/2009
1 12/6/2009
2 09/3/2009
2 12/6/2009
3 09/3/2009
...
In that case, you could find out how many days off each person took by using the following query:
SELECT EMPLOYEE_ID, COUNT(*) AS NUM_DAYS_OFF FROM DAYS_OFF_TABLE GROUP BY EMPLOYEE_ID
And the number of people who took days off on each date like this:
SELECT DAY_OFF, COUNT(*) AS NUM_PEOPLE FROM DAYS_OFF_TABLE GROUP BY DAY_OFF
But I digress...
You can try to use an SQL CASE statement to help with this:
SELECT Employee, CASE
WHEN First_Day_Off is NULL AND Second_Day_Off is NULL THEN 0
WHEN First_Day_Off is NOT NULL AND Second_Day_Off is NULL THEN 1
WHEN First_Day_Off is NULL AND Second_Day_Off is NOT NULL THEN 1
ELSE 2
END AS NUM_DAYS_OFF
FROM DAYS_OFF_TABLE
(note that you may need to change around the syntax slightly depending on your database.
Getting dates and number of people who took off on that day might be more complicated.
I don't know if this would work, but you can try it:
SELECT
Date_Off,
COUNT(*) AS Num_People
FROM
(SELECT
First_Day_Off, COUNT(*) AS Num_People FROM DAYS_OFF_TABLE WHERE First_Day_Off IS NOT NULL GROUP BY First_Day_Off
UNION
SELECT Second_Day_Off, COUNT(*) AS Num_People FROM DAYS_OFF_TABLE WHERE Second_Day_Off IS NOT NULL GROUP BY Second_Day_Off)
GROUP BY
Num_People