I am trying to find whether a person (id = A3) is continuously active in a program at least five months or more in a given year (2013). Any suggestion would be appreciated. My data look like as follows:
You simply use group by and a conditional expression:
select id,
(case when count(ActiveMonthYear) >= 5 then 'YES!' else 'NAW' end)
from table t
where ListOfTheMonths between '201301' and '201312'
group by id;
EDIT:
I suppose "continuously" doesn't just mean any five months. For that, there are various ways. I like the difference of row numbers approach
select distinct id
from (select t.*,
(row_number() over (partition by id order by ListOfTheMonths) -
count(ActiveMonthYear) over (partition by id order by ListOfTheMonths)
) as grp
from table t
where ListOfTheMonths between '201301' and '201312'
) t
where ActiveMonthYear is not null
group by id, grp
having count(*) >= 5;
The difference in the subquery is constant for groups of consecutive active months. This is then used a grouping. The result is a list of all ids that meet this criteria. You can add a where for a particular id (do it in the subquery).
By the way, this is written using select distinct and group by. This is one of the rare cases where these two are appropriately used together. A single id could have two periods of five months in the same year. There is no reason to include that person twice in the result set.
Related
I would like to get a new ID, no matter the format (in the example below 11,12,13...)
Based on the following condition:
Every time the days column value is greater then 1 and not null then current row and all following ones will get the same ID until a new value will meet the condition.
Within the same email
Below you can see the expected 1 (in the format of XX)
I thought about using two conditions with the following order between them
Every time the days column value is greater then 1 then all following rows will get the same ID until a new value will meet the condition.
2.AND When lag (previous) is equal to 0/1/null.
Assuming you have an EmailDate column over which you're ordering (a DATETIME field, really), try something like this:
WITH
TableNameWithEmailDateIDs AS (
SELECT
*,
ROW_NUMBER() OVER (
ORDER BY
Email DESC,
EmailDate
) AS EmailDateID
FROM
TableName
),
IDs AS (
SELECT
*,
LEAD(EmailDateID, 1) OVER (
ORDER BY
Email,
EmailDate
) AS LeadEmailDateID
FROM
(
SELECT
*,
-- REMOVE +10 if you don't want 11 to be starting ID
ROW_NUMBER() OVER (
ORDER BY
Email DESC,
EmailDate
)+10 AS ID
FROM
TableNameWithEmailDateIDs
WHERE
Days > 1
OR Days IS NULL
) X
)
SELECT
COALESCE(TableName.EmailDate, IDs.EmailDate) AS EmailDate,
IDs.Email,
COALESCE(TableName.Days, IDs.Days) AS Days,
IDs.ID
FROM
IDs
LEFT JOIN TableNameWithEmailDateIDs TableName
ON IDs.Email = TableName.Email
AND TableName.EmailDateID BETWEEN
IDs.EmailDateID
AND IDs.LeadEmailDateID-1
ORDER BY
ID DESC,
TableName.EmailDate DESC
;
First, create a CTE that generates IDs for each distinct Email/Date combo (helpful for LEFT JOIN condition later). Then, create a CTE that generates IDs for rows that meet your condition (i.e. the important rows). Finally, LEFT JOIN your main table onto that CTE to fill in the "gaps", so to speak.
I suggest running each of the components of this query independently to fully understand what's going on.
Hope it helps!
Recently, i got a table which name Appointments
The requirement is that i need to select only one row for each customer by 2 rule:
if same time and (same location or different location), put null on tutor and location.
if different time and (same location or different location), pick the smallest row.
Since i'm so amateur in SQL, i've search the method of self join, but it seems not working in this case.
Expected result
Thanks all, have a great day...
You seem to want the minimum time for each customer, with null values if there are multiple rows and the tutor or location don't match.
You can use window functions:
select customer, starttime,
(case when min(location) = max(location) then min(location) end) as location,
(case when min(tutor) = max(tutor) then min(tutor) end) as tutor
from (select t.*, rank() over (partition by customer order by starttime) as seqnum
from t
) t
where seqnum = 1
group by customer, starttime
I'm a new SQL user and need help.
Let's say I have a vehicle number 123 and I've traveled from Region 3 to final destination Region 4. In between, I've visited Region 1 and 5 as well but that's not my concern.
Simple example would be as follow.
Original Table
Desired Output
How can this be done in SQL query?
You have a sequence number so you can use some form of aggregation. One method is:
select records,
max(case when sequence = 1 then fromregion end) as fromregion,
max(case when sequence = maxsequence then toregion) as toregion
from (select t.*, max(sequence) over (partition by records) as max_sequence
from t
) t
group by records;
Unfortunately, SQL Server doesn't offer "first()" or "last()" as aggregation functions. But it does support first_value() as a window function. This allows you to do the logic without a subquery:
select distinct records,
first_value(fromRegion) over (partition by records order by sequence) as fromregion,
first_value(toRegion) over (partition by records order by sequence desc) as toregion
from t;
I have a table 'Cashup_Till' that records all data on what a particular till has recorded in a venue for a given day, each venue has multiple tills all with a designated number 'Till_No'. I need to get the previous 2 days entries for each till number. For each till Individually I can do this...
SELECT TOP 2 T.* FROM CashUp_Till T
WHERE T.Till_No = (Enter Till Number Here)
ORDER BY T.Till_Id DESC
Some venues have 20-30 tills so Ideally I need to do all the tills in one call. I can pass in a user defined table type of till numbers, then select them in a subquery, but that's as far as my SQL knowledge takes me, does anyone have a solution?
Here is one way:
SELECT T.*
FROM (SELECT T.*,
ROW_NUMBER() OVER (PARTITION BY Till_No ORDER BY Till_Id DESC) as seqnum
FROM CashUp_Till T
) T
WHERE seqnum <= 2;
This assumes that there is one record per day, which I believe is suggested by the question.
If you have a separate table of tills, then:
select ct.*
from t cross apply
(select top 2 ct.*
from cashup_till ct
where ct.till_no = t.till_no
order by till_id desc
) ct;
I have a table - Data - of rows, simplified, like so:
Name,Amount,Last,Date
A,16,31,1-Jan-2014
A,27,38,1-Feb-2014
A,12,34,1-Mar-2014
B,8,37,1-Jan-2014
B,3,38,1-Feb-2014
B,17,39,1-Mar-2014
I wish to group them similar to:
select Name,sum(Amount),aggr(Last),max(Date) from Data group by Name
For aggr(Last) I want the value of 'Last' from the row that contains max(Date)
So the result I want would be 2 rows
Name,Amount,Last,Date
A,55,34,1-Mar-2014
B,28,39,1-Mar-2014
i.e. in both cases, the value of Last is the one from the row that contained 1-Mar-2014
The query I'm actually doing is basically the same, but with many more sum() fields and millions of rows, so I'm guessing an aggregate function could avoid multiple extra requests each group of incoming rows.
Instead, use row_number() and conditional aggregation:
select Name, sum(Amount),
max(case when seqnum = 1 then Last end) as Last,
max(date)
from (select d.*, row_number() over (partition by name order by date desc) as seqnum
from data d
) d
group by Name;