I need to get first and last record (ordered by Date column) from table for certain SSID. It is not a problem if there is more records with same max or min date. All I need is union all.
I am getting last record having max(date) with:
with c as (
select *, rnk = rank() over (partition by Date order by Date ASC)
from table
where SSID = '00921834800'
)
select top 1 Date, City, Title
from c
order by Date desc
How to I get first record (min(Date)) as well (same thing only with order by Date asc) with single select and without using ranking again?
I'm using MSSQL 2017.
; with c as (
select *,
rnk = rank() over (partition by Date order by Date ASC),
rnk2 = rank() over (partition by Date order by Date desc)
from table
where SSID= '00921834800'
)
select Date,
City,
Title
from c
where rnk = 1 or rnk2 = 1
order by Date desc
I would use the following query:
select * from (select top 1 with ties * from t where ssid = '00921834800' order by date) as a
union all
select * from (select top 1 with ties * from t where ssid = '00921834800' order by date desc) as b
One other solution is :
with
c as
(
select *,
rank() over (partition by Date order by Date ASC) AS RNK,
count() OVER (partition by Date) AS CNT
from table
where SSID= '00921834800')
select Date, City, Title
from c
WHERE RNK = 1
OR CNT = RNK
order by Date desc
Related
I have a table with Name and Date columns. I want to get the earliest date when the current name appeared. For example:
Name
Date
X
30-Jan-2021
X
29-Jan-2021
X
28-Jan-2021
Y
27-Jan-2021
Y
26-Jan-2021
Y
25-Jan-2021
Y
24-Jan-2021
X
23-Jan-2021
X
22-Jan-2021
Now when I try to get the earliest date when current name (X) started to appear, I want 28-Jan, but the sql query would give 22-Jan-2021 because that's when X appeared originally for the first time.
Update: This was the query I was using:
Select min(Date) from myTable where Name='X'
I am using older SQL Server 2008 (in the process of upgrading), so do not have access to LEAD/LAG functions.
The solutions suggested below do work as intended. Thanks.
This is a type of gaps-and-islands problem.
There are many solutions. Here is one that is optimized for your case
Use LEAD/LAG to identify the first row in each grouping
Filter to only those rows
Number them rows and take the first one
WITH StartPoints AS (
SELECT *,
IsStart = CASE WHEN Name <> LEAD(Name, 1, '') OVER (ORDER BY Date DESC) THEN 1 END
FROM YourTable
),
Numbered AS (
SELECT *,
rn = ROW_NUMBER() OVER (PARTITION BY Name ORDER BY Date DESC)
FROM StartPoints
WHERE IsStart = 1 AND Name = 'X'
)
SELECT
Name, Date
FROM Numbered
WHERE rn = 1;
db<>fiddle
For SQL Server 2008 or earlier (which I strongly suggest you upgrade from), you can use a self-join with row-numbering to simulate LEAD/LAG
WITH RowNumbered AS (
SELECT *,
AllRn = ROW_NUMBER() OVER (ORDER BY Date ASC)
FROM YourTable
),
StartPoints AS (
SELECT r1.*,
IsStart = CASE WHEN r1.Name <> ISNULL(r2.Name, '') THEN 1 END
FROM RowNumbered r1
LEFT JOIN RowNumbered r2 ON r2.AllRn = r1.AllRn - 1
),
Numbered AS (
SELECT *,
rn = ROW_NUMBER() OVER (PARTITION BY Name ORDER BY Date DESC)
FROM StartPoints
WHERE IsStart = 1
)
SELECT
Name, Date
FROM Numbered
WHERE rn = 1;
This is a gaps and island problem. Based on the sample data, this will work:
WITH Groups AS(
SELECT YT.[Name],
YT.[Date],
ROW_NUMBER() OVER (ORDER BY YT.Date DESC) -
ROW_NUMBER() OVER (PARTITION BY YT.[Name] ORDER BY Date DESC) AS Grp
FROM dbo.YourTable YT),
FirstGroup AS(
SELECT TOP (1) WITH TIES
G.[Name],
G.[Date]
FROM Groups G
WHERE [Name] = 'X'
ORDER BY Grp ASC)
SELECT MIN(FG.[Date]) AS Mi
db<>fiddle
If i did understand, you want to know when the X disappeared and reappeared again. in that case you can search for gaps in dates by group.
this and example how to detect that
SELECT name
,DATE
FROM (
SELECT *
,DATEDIFF(day, lead(DATE) OVER (
PARTITION BY name ORDER BY DATE DESC
), DATE) DIF
FROM YourTable
) a
WHERE DIF > 1
I need to select the data of all my customers with the records displayed in the image. But I need to get the most recent record only, for example I need to get the order # E987 for John and E888 for Adam. As you can see from the example, when I do the select statement, I get all the order records.
You don't mention the specific database, so I'll answer with a generic solution.
You can do:
select *
from (
select t.*,
row_number() over(partition by name order by order_date desc) as rn
from t
) x
where rn = 1
You can use analytical function row_number.
Select * from
(Select t.*,
Row_number() over (partition by customer_id order by order_date desc) as rn
From your_table t) t
Where rn = 1
Or you can use not exists as follows:
Select *
From yoir_table t
Where not exists
(Select 1 from your_table tt
Where t.customer_id = tt.custome_id
And tt.order_date > t.order_date)
You can do it with a subquery that finds the last order date.
SELECT t.*
FROM yoir_table t
JOIN (SELECT tt.custome_id,
MAX(tt.order_date) MaxOrderDate
FROM yoir_table tt
GROUP BY tt.custome_id) AS tt
ON t.custome_id = tt.custome_id
AND t.order_date = tt.MaxOrderDate
I am currently getting the output of the image below. I want to be able to retrieve the latest Turn Time. Essentially the MAX beginning date and MAX end date. How Should I structure my query ?
I think you just want order by:
select top (1) t.*
from t
order by enddate desc, beginning_date desc;
If you want this per id, then you can use window functions or top (1) with ties:
select top (1) t.*
from (select t.*,
row_number() over (partition by id order by enddate desc, beginning_date desc) as seqnum
from t
) t
where seqnum = 1;
You can use row_number()
select * from
(
select *,row_number() over(parition by id order by beginningdate desc) as rn
from tablename
)A where rn=1
For the larger turn time -
select * from
(
select *,row_number() over(parition by id order by turntime desc) as rn
from tablename
)A where rn=1
I have the following table:
I want to get the most recent status for each dept_code that a CL_ID has. So the desired output would be this:
I have tried the following but this give me just the most recent status for each client and not each of their dept_codes.
SELECT *
FROM [CIMSHR6_MERGED].[dbo].[C3CLSTAT] C
INNER JOIN
(SELECT CLIENT_NUMBER, MAX(STATUS_DATE) AS SDATE
FROM [CIMSHR6_MERGED].[dbo].[C3CLSTAT]
GROUP BY CLIENT_NUMBER) X
ON X.CLIENT_NUMBER = C.CLIENT_NUMBER
AND X.SDATE = C.STATUS_DATE
ORDER BY C.CLIENT_NUMBER
Any help would be much appreciated. Thanks.
A convenient method that works in SQL Server is:
select top (1) cl.*
from [CIMSHR6_MERGED].[dbo].[C3CLSTAT] cl
order by row_number() over (partition by cl_id, dept_code order by status_date desc);
A method that is efficient with the right indexes in almost any database is:
select cl.*
from [CIMSHR6_MERGED].[dbo].[C3CLSTAT] cl
where cl.status_date = (select max(cl2.status_date)
from [CIMSHR6_MERGED].[dbo].[C3CLSTAT] cl2
where cl2.cl_id = cl.cl_id and cl2.dept_code = cl.dept_code
);
The right index is on (cl_id, dept_code, status_date).
I would also use ROW_NUMBER, but with a subquery:
SELECT CL_ID, Status_date, Status, Dept_code
FROM
(
SELECT *,
ROW_NUMBER() OVER (PARTITION BY CL_ID, Dept_code ORDER BY Status_date DESC) rn
FROM CIMSHR6_MERGED].[dbo].[C3CLSTAT]
) t
WHERE rn = 1;
1) Firstly group everything on Dept_Code,CL_ID and assign rank for each row with in the group in descending order.
2) Select all the rows with rnk=1 which would display your desired result.
SELECT Z.CL_ID,
Z.Status_Date,
Z.Status,
Z.Dept_Code
FROM
(
SELECT *,
RANK() OVER( PARTITION BY Dept_Code,CL_ID, ORDER BY Status_Date DESC ) AS rnk
FROM [CIMSHR6_MERGED].[dbo].[C3CLSTAT]
) Z
WHERE Z.rnk = 1;
This would work for almost all databases
select * from c3clstat c
where exists
(select 1 from c3clstat c1
where c1.cl_id=c.cl_id
and c1.dept_code=c.dept_code
group by cl_id,dept_code
having c.status_date=max(c1.status_date)
)
I have a table with below mentioned columns. I want to fetch the previous status of customer. Once customer id can have multiple entries
Customer_id status start_date end_date Active
1 Member 01-JAN-18 04-FEB-18 N
1 Explorist 05-FEB-18 30-APR-18 N
1 Globalist 01-MAY-18 31-DEC-99 Y
Desired output
Customer _id Previous_status end_date
1 Explorist 30-APR-18
Please try below query using QUALIFY keyword and ROW_NUMBER():
SELECT a.* from table a
QUALIFY ROW_NUMBER OVER(PARTITION BY customer_id order by start_date desc) = 2
Below query should work.
SELECT * from (
SELECT a.*,
ROW_NUMBER() over (partition by customer_id order by start_date desc) rn
from table a )
where rn =2
You can use below query and I guess that is very simple and that worked for me,
select * from customer order by end_date desc limit 1,1
Consider this question: Select Nth Row From A Table In Oracle
In your case, that would be:
select * from (select a.*, rownum rnum from (select * from <your table name>
order by <start_date or end_date> desc) a where rownum <= 2) where rnum >= 2;
If you are using Oracle DataBase then try below query using ROW_NUMBER() function:Let's consider the table name is customer
SELECT TEMP.CUSTOMER_ID
,TEMP.STATUS
,TEMP.START_DATE
,TEMP.END_DATE
,TEMP.ACTIVE
FROM(
SELECT ROW_NUMBER() OVER (PARTITION BY CUSTOMER_ID ORDER BY CUSTOMER_ID ASC,START_DATE DESC) AS "ROW_NUM"
,CUSTOMER_ID
,STATUS
,START_DATE
,END_DATE
,ACTIVE
FROM CUSTOMER) TEMP
WHERE TEMP."ROW_NUM" = 2;