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;
Related
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
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 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)
)
How to fetch records with latest 2 version numbers from a table using SQL query.
I want to fetch user ids for a tracking id for the last 2 versions.
Below is my table description:
TRACKING_ID ,
User_id,
Version_number
Below query gives me the user id having the latest version.
Select user_id
from table t1
join
(select tracking_id,max(version_number) as version_number
from table
group by tracking_id ) t2
on t1.tracking_id=t2.tracking_id
and t1.version_number=t2.version_number
Appreciate your response.
You can use ROW_NUMBER:
SELECT tracking_id,
user_id,
version_number
FROM ( SELECT *,
ROW_NUMBER() OVER(PARTITION BY tracking_id ORDER BY version_number DESC) AS RN
FROM YourTable) AS T
WHERE RN <= 2
I don't fully understand the question but something like this may point you in the correct direction.
WITH CTE AS (
SELECT tracking_ID, user_ID, version_number,
Row_number() over (partition by USER_ID, Version_number desc) as RN)
SELECT *
FROM cte
WHERE RN <= 2;
I'm not sure if the user_ID or the tracking_ID needs to be partitioned...
My requirement is to get each client's latest order, and then get top 100 records.
I wrote one query as below to get latest orders for each client. Internal query works fine. But I don't know how to get first 100 based on the results.
SELECT * FROM (
SELECT id, client_id, ROW_NUMBER() OVER(PARTITION BY client_id ORDER BY create_time DESC) rn
FROM order
) WHERE rn=1
Any ideas? Thanks.
Assuming that create_time contains the time the order was created, and you want the 100 clients with the latest orders, you can:
add the create_time in your innermost query
order the results of your outer query by the create_time desc
add an outermost query that filters the first 100 rows using ROWNUM
Query:
SELECT * FROM (
SELECT * FROM (
SELECT
id,
client_id,
create_time,
ROW_NUMBER() OVER(PARTITION BY client_id ORDER BY create_time DESC) rn
FROM order
)
WHERE rn=1
ORDER BY create_time desc
) WHERE rownum <= 100
UPDATE for Oracle 12c
With release 12.1, Oracle introduced "real" Top-N queries. Using the new FETCH FIRST... syntax, you can also use:
SELECT * FROM (
SELECT
id,
client_id,
create_time,
ROW_NUMBER() OVER(PARTITION BY client_id ORDER BY create_time DESC) rn
FROM order
)
WHERE rn = 1
ORDER BY create_time desc
FETCH FIRST 100 ROWS ONLY)
you should use rownum in oracle to do what you seek
where rownum <= 100
see also those answers to help you
limit in oracle
select top in oracle
select top in oracle 2
As Moneer Kamal said, you can do that simply:
SELECT id, client_id FROM order
WHERE rownum <= 100
ORDER BY create_time DESC;
Notice that the ordering is done after getting the 100 row. This might be useful for who does not want ordering.
Update:
To use order by with rownum you have to write something like this:
SELECT * from (SELECT id, client_id FROM order ORDER BY create_time DESC) WHERE rownum <= 100;
First 10 customers inserted into db (table customers):
select * from customers where customer_id <=
(select min(customer_id)+10 from customers)
Last 10 customers inserted into db (table customers):
select * from customers where customer_id >=
(select max(customer_id)-10 from customers)
Hope this helps....
To select top n rows updated recently
SELECT *
FROM (
SELECT *
FROM table
ORDER BY UpdateDateTime DESC
)
WHERE ROWNUM < 101;
Try this:
SELECT *
FROM (SELECT * FROM (
SELECT
id,
client_id,
create_time,
ROW_NUMBER() OVER(PARTITION BY client_id ORDER BY create_time DESC) rn
FROM order
)
WHERE rn=1
ORDER BY create_time desc) alias_name
WHERE rownum <= 100
ORDER BY rownum;
Or TOP:
SELECT TOP 2 * FROM Customers; //But not supported in Oracle
NOTE: I suppose that your internal query is fine. Please share your output of this.