I Want to extract those account ids which are present in may and JUNE but not in JULY . I have data for 6 months
i have tried using CONDITION(AND/ NOT)
SELECT ACCOUNT_ID, DATA_MONTH
FROM DATA1
WHERE DATA_MONTH ='01-MAY-18' AND DATA_MONTH <> '01-JULY-18
GROUP BY DATA_MONTH
I am still getting account ids which where in JULY
You can put the conditions you need in a HAVING clause:
SELECT ACCOUNT_ID
FROM DATA1
GROUP BY ACCOUNT_ID
HAVING
SUM(CASE WHEN MONTH(DATA_MONTH) = 5 THEN 1 ELSE 0 END) > 0
AND
SUM(CASE WHEN MONTH(DATA_MONTH) = 6 THEN 1 ELSE 0 END) > 0
AND
SUM(CASE WHEN MONTH(DATA_MONTH) = 7 THEN 1 ELSE 0 END) = 0
I assume that in your rdbms there exists a function like MONTH() to extract the month from the date.
If you want to check the year also, you can do it in each condition, so you must change to:
CASE WHEN MONTH(DATA_MONTH) = 7 AND YEAR(DATA_MONTH) = 2018 THEN 1 ELSE 0
Using SQL Server's Month function gives you an understandable example of how to solve this.
SELECT ACCOUNT_ID
FROM DATA1
WHERE MONTH(DATA_MONTH) in (5,6) AND YEAR(DATA_MONTH) = 2018 AND MONTH(DATA_MONTH) NOT IN (7)
GROUP BY ACCOUNT_ID
Your date formats look like Oracle, so I will answer for that database.
The idea is to use conditional aggregation:
SELECT ACCOUNT_ID, DATA_MONTH
FROM DATA1
WHERE DATA_MONTH >= DATE '2018-05-01' AND
DATA_MONTH < DATE '2018-08-01'
GROUP BY ACCOUNT_ID
HAVING COUNT(DISTINCT DATA_MONTH) = 2 AND
MAX(DATA_MONTH) < DATE '2018-07-01';
This assumes that all your dates are on the first of the month.
SELECT ACCOUNT_ID, DATA_MONTH
FROM DATA1
WHERE MONTH(DATA_MONTH) =5 AND MONTH(DATA_MONTH) = 6
GROUP BY DATA_MONTH
SELECT ACCOUNT_ID, DATA_MONTH
FROM DATA1
WHERE MONTH(DATA_MONTH) BETWEEN 5 AND 6
GROUP BY Year(DATA_MONTH),month(DATA_MONTH)
Related
First here's a sample table.
enter image description here
Provider_name patient date status length
AF AGUIR00001 07/05/2018 3 30
AF ABBOT00001 07/05/2018 30
BB ADAMS00001 07/05/2018 3 30
BB ACEVE00001 07/06/2018 3 30
I have created a query that lets me count the total number of appointments versus the number of appointments with a certain status(eg checked out). I was able to create it and group it by provider.
select provider_name,
count(patient) total,
sum(case when status = 3 then 1 else 0 end) as Checkedout
from appointment
group by provider_name
Then I moved on to the next phase which was to get the total length of those appointments with checkedout status. I made this query but it does not break down into each provider.
select provider_name,
count(patient) total,
sum(case when status = 3 then 1 else 0 end) as Checkedout,
(select sum(length) from appointment where status = 3
and date between '06/01/2018' and '07/06/2018')
from appointment where date between '06/01/2018' and '07/06/2018'
group by provider_name
I need it so that the last column in the query is segregated per provider_name.
Thank you in advance for helping me out.
Actually, you were on the right way, try this:
select provider_name,
count(patient) total,
sum(case when status = 3 then 1 else 0 end) as Checkedout,
sum(case when status = 3 then length else 0 end) as len_status3
from appointment
where date between '2018-01-06' and '2018-06-07'
group by provider_name;
According to your last comment, you need a WITH ROLLUP modifier for GROUP BY as in the following :
select coalesce(provider_name,'Total') as provider_name,
count(patient) total,
sum(case when status = 3 then 1 else 0 end) as Checkedout,
sum(case when status = 3 then length else 0 end) as len_status3
from appointment
where date between '2018-01-06' and '2018-06-07'
group by provider_name with rollup;
SQL Fiddle Demo
you shoul do as for checkedoutout
select provider_name,
count(patient) total,
sum(case when status = 3 then 1 else 0 end) as Checkedout,
sum( case when status = 3 then length else 0 ) as total_length
from appointment where date between '06/01/2018' and '07/06/2018'
group by provider_name
How can I get the count per hour?
select count_hr_1, count_hr_2, count_hr_3 from db.table where year=2018 and month=01 and day=02 and hour=01 OR hour=02 OR hour=03;
This SQL/Query is probably invalid but I want to get the counts of hour 1,2,3
If you are able to process the resultset as multiple rows instead of one, you could use GROUP BY:
select hour, count(*)
from db.table
where
year = 2018
and month = 1
and day = 2
and hour in (1, 2, 3)
group by hour
select sum(case when hour = 1 then 1 else 0 end) as count_hr_1,
sum(case when hour = 2 then 1 else 0 end) as count_hr_2,
sum(case when hour = 3 then 1 else 0 end) as count_hr_3
from db.table
where year = 2018
and month = 1
and day = 2
and hour in (1,2,3)
I got a table of Visitors.
Visitor has the following columns:
Id
StartTime (Date)
Purchased (bool)
Shipped (bool)
For each day within the last 7 days, I want to select 3 counts of the Visitors who have that day as StartTime:
The count of total visitors
The count of total visitors where Purchased = true
The count of total visitors where Shipped = true
Ideally the returned result would be:
Day Total TotalPurchased TotalShipped
1 100 67 42
2 82 61 27
etc...
I am used to .NET Linq so this has proved to be quite a challenge for me.
All I have come up with so far is the following:
SELECT COUNT(*) AS Total
FROM [dbo].[Visitors]
WHERE DAY([StartTime]) = DAY(GETDATE())
It selects the total of the current day just fine, however I feel pretty stuck right now so it'd be nice if someone could point me in the right direction.
For the last 7 days use the query proposed by Stanislav but with this WHERE clause
SELECT DAY([StartTime]) theDay,
COUNT(*) AS Tot,
SUM(CASE WHEN Purchased=true THEN 1 ELSE 0 END) as TotPurch,
SUM(CASE WHEN Shipped=true THEN 1 ELSE 0 END) as TotShip
FROM [dbo].[Visitors]
WHERE [StartTime] BETWEEN GETDATE()-7 AND GETDATE()
GROUP BY DAY([StartTime])
SELECT COUNT(*) AS Total,
SUM(CASE WHEN Purchased=true THEN 1 ELSE 0 END) as TotalPurchased,
SUM(CASE WHEN Shipped=true THEN 1 ELSE 0 END) as TotalShipped
FROM [dbo].[Visitors]
WHERE DAY([StartTime]) = DAY(GETDATE())
and add GROUP BY DAY([StartTime]) as jarlh mentioned
Here's a simple select that will give you the dataset you want
SELECT DATEDIFF(day,StartTime, getdate())+1 as [Day], -- Add 1 to display 1 to 7 instead of 0 to 6
COUNT(*) as Total,
SUM(CASE WHEN Purchased = 1 THEN 1 ELSE 0 END) as TotalPurchased,
SUM(CASE WHEN Shipped = 1 THEN 1 ELSE 0 END) AS TotalShipped
FROM Visitors
WHERE DATEDIFF(day,startTime,GETDATE()) < 6
GROUP BY DATEDIFF(day,startTime,GETDATE())
ORDER BY 1
This query will not take into consideration the time component of the date.
I'm sorry if the title wasn't too clear, but the following explanation will be more accurate.
I have the following view:
DATE USER CONDITION
20140101 1 A
20140101 2 B
20140101 3 C
20140108 1 C
20140108 3 B
20140108 2 C
What I need to do is present how many users where in all conditions this week and 7 days before today.
Output should be like this:
Condition Today Last_Week (Today-7)
A 0 1
B 1 1
C 2 1
How can I do this in Oracle? I will need to do this for 4 weeks so itll be Today-7,14-21.
I've tried this with group by but I get the "week2" as rows. Then I've tried something like Select conditions, (select count(users) from MyView where DATE='Today') FROM MyView(looking at something thats actually working) but it doesnt work for me.
Achieved this with a little modification of the accepted answer:
select condition,
count(case when to_date(xdate) = to_date(sysdate) then 1 end) to_day,
count(case when to_date(xdate) = to_date(sysdate-7) then 1 end) last_7_days
from my_table
group by condition
select condition, count(case when to_date(xdate) = to_date(sysdate) then 1 end) to_day,
count(case when to_date(xdate) < to_date(sysdate) then 1 end) last_7_days
from my_table
where to_date(xdate) >= to_date(sysdate) - 7
group by condition
select condition
, sum
( case
when date between trunc(sysdate) - 7 and trunc(sysdate) - 1
then 1
else 0
end
)
last_week
, sum
( case
when date between trunc(sysdate) and trunc(sysdate + 1)
then 1
else 0
end
)
this_week
from table
group
by condition
By using the conditional count (as a sum) and grouping on condition you can filter out all desired dates. Note that using trunc will cause to use the begin of the day.
I have a simple SQL table of the following format:
date | value
============
2010-01-01 | -54
2010-01-02 | 134
2010-01-03 | 73
With the following two SQL queries I can select the sum of values and distinguish between the sum of positive values and the sum of negative values.
SELECT YEAR(date), SUM(value) FROM table WHERE value > 0 GROUP BY YEAR(date)
SELECT YEAR(date), SUM(value) FROM table WHERE value < 0 GROUP BY YEAR(date)
Is there a way to do this in a single SQL query?
Thanks in advance!
Not tested (don't know if the case can be put within a SUM instruction), but the CASE instruction could be of help:
SELECT
YEAR(date),
SUM(CASE WHEN value > 0 THEN value ELSE 0 END) as positives_sum,
SUM(CASE WHEN value < 0 THEN value ELSE 0 END) as negatives_sum
FROM
table
GROUP BY
YEAR(date)
Many databases have a SIGN() function
SELECT YEAR(date),
SUM(value)
FROM table
GROUP BY YEAR(date),
SIGN(value)