For a single id column, we have sequence number 01, 02, 03 upto 99 repeating twice /thrice.
Example:
ID SEQ_NO
----------
2 01
2 02
2 03
.
.
.
2 99
2 01
2 02
2 99
We have a requirement to add AA prefix to second time when it is looping on seq_no, and for third time it should be BB.
Can anyone explain how to do this?
Try the following using the ROW_NUMBER function:
If you want only to select SEQ_NO as a new column:
WITH CTE AS
(
SELECT *, ROW_NUMBER() OVER (PARTITION BY ID, SEQ_NO ORDER BY SEQ_NO) rn
FROM table_name
)
SELECT ID, SEQ_NO,
CASE
WHEN rn>1 THEN
CONCAT(CHAR(rn+63), CHAR(rn+63), SEQ_NO)
ELSE SEQ_NO
END AS new_seq
FROM CTE
WHERE rn <= 27
ORDER BY ID, new_seq
If you want to update the SEQ_NO column:
WITH CTE AS
(
SELECT *, ROW_NUMBER() OVER (PARTITION BY ID, SEQ_NO ORDER BY SEQ_NO) rn
FROM table_name
)
UPDATE CTE SET SEQ_NO = CONCAT(CHAR(rn+63), CHAR(rn+63), SEQ_NO)
WHERE rn > 1 AND rn <= 27
See a demo with a set of data where seq (01 - 10) is repeated three times.
Related
Is it possible to rank item by partition without use CTE method
Expected Table
item
value
ID
A
10
1
A
20
1
B
30
2
B
40
2
C
50
3
C
60
3
A
70
4
A
80
4
By giving id to the partition to allow agitated function to work the way I want.
item
MIN
MAX
ID
A
10
20
1
B
30
40
2
C
50
60
3
A
70
80
4
SQL Version: Microsoft SQL Sever 2017
Assuming that the value column provides the intended ordering of the records which we see in your question above, we can try using the difference in row numbers method here. Your problem is a type of gaps and islands problem.
WITH cte AS (
SELECT *, ROW_NUMBER() OVER (ORDER BY value) rn1,
ROW_NUMBER() OVER (PARTITION BY item ORDER BY value) rn2
FROM yourTable
)
SELECT item, MIN(value) AS [MIN], MAX(value) AS [MAX], MIN(ID) AS ID
FROM cte
GROUP BY item, rn1 - rn2
ORDER BY MIN(value);
Demo
If you don't want to use a CTE here, for whatever reason, you may simply inline the SQL code in the CTE into the bottom query, as a subquery:
SELECT item, MIN(value) AS [MIN], MAX(value) AS [MAX], MIN(ID) AS ID
FROM
(
SELECT *, ROW_NUMBER() OVER (ORDER BY value) rn1,
ROW_NUMBER() OVER (PARTITION BY item ORDER BY value) rn2
FROM yourTable
) t
GROUP BY item, rn1 - rn2
ORDER BY MIN(value);
You can generate group IDs by analyzing the previous row item value that could be obtained with the LAG function and finally use GROUP BY to get the minimum and maximum value in item groups.
SELECT
item,
MIN(value) AS "min",
MAX(value) AS "max",
group_id + 1 AS id
FROM (
SELECT
*,
SUM(CASE WHEN item = prev_item THEN 0 ELSE 1 END) OVER (ORDER BY value) AS group_id
FROM (
SELECT
*,
LAG(item, 1, item) OVER (ORDER BY value) AS prev_item
FROM t
) items
) groups
GROUP BY item, group_id
Query produces output
item
min
max
id
A
10
20
1
B
30
40
2
C
50
60
3
A
70
80
4
You can check a working demo here
I have records like this and I want to search in SQL Server by pattern and sequence like (50,54,50) in value field it should return 02,03,04 any one have idea to do this.
======================================
Id Date Value
01 2020-01-01 50
02 2020-01-02 50
03 2020-01-03 54
04 2020-01-04 50
05 2020-01-05 35
06 2020-01-06 98
07 2020-01-07 13
======================================
There is a request on the user voice site Add support for Row Pattern Recognition in T-SQL (SQL:2016 features R010 and R020) which I believe would allow for this.
In the meantime this should do what you need
WITH T AS
(
SELECT *,
LAG(Id) OVER (ORDER BY Id) AS PrevId,
LAG(value) OVER (ORDER BY Id) AS PrevValue,
LEAD(Id) OVER (ORDER BY Id) AS NextId,
LEAD(value) OVER (ORDER BY Id) AS NextValue
FROM YourTable
)
SELECT PrevId, Id, NextId
FROM T
WHERE PrevValue = 50 AND Value =54 AND NextValue = 50
If you wanted a more flexible approach, you can use cross apply:
select t2.*
from t cross apply
(select string_agg(id, ',') within group (order by date) as ids,
string_agg(value, ',') within group (order by date) as vals
from (select top (3) t2.*
from t t2
where t2.date >= t.date
order by t2.date
) t2
) t2
where vals = '50,54,50';
Here is a db<>fiddle.
If string_agg() were supported as a window function, you could use:
select t.*
from (select t.*,
string_agg(id, ',') within group (order by date) over (order by id rows between current row and 2 following) as ids,
string_agg(value, ',') within group (order by date) over (order by id rows between current row and 2 following) as vals
from t
) t
where vals = '50,54,50';
But alas, it is not.
If I get your requirement correct, yo can try this below logic developed with the help of LAG and LEAD-
DEMO HERE
WITH CTE
AS
(
SELECT Id,Date,
LAG(value,2) OVER(ORDER BY id) lag_2,
LAG(value,1) OVER(ORDER BY id) lag_1,
Value c_val,
LEAD(value,1) OVER(ORDER BY id) lead_1,
LEAD(value,2) OVER(ORDER BY id) lead_2
FROM your_table
)
SELECT Id,Date,
CASE
WHEN (lag_2 = 50 AND lag_1 = 54 AND c_val = 50) OR
(lag_1 = 50 AND c_val = 54 AND lead_1 = 50) OR
(c_val = 50 AND lead_1 = 54 AND lead_2 = 50)
THEN (
CASE
WHEN lead_1 = 54 THEN 02
WHEN c_val = 54 THEN 03
WHEN lag_1 = 54 THEN 04
END
)
ELSE c_val
END
FROM CTE
I have a table that has has some measurements, ID and date.
The table is built like so
ID DATE M1 M2
1 2020 1 NULL
1 2020 NULL 15
1 2018 2 NULL
2 2019 1 NULL
2 2019 NULL 1
I would like to end up with a table that has one row per ID with the most recent measurement
ID M1 M2
1 1 15
2 1 1
Any ideas?
You can use correlated sub-query with aggregation :
select id, max(m1), max(m2)
from t
where t.date = (select max(t1.date) from t t1 where t1.id = t.id)
group by id;
Use ROW_NUMBER combined with an aggregation:
WITH cte AS (
SELECT *, ROW_NUMBER() OVER (PARTITION BY ID ORDER BY DATE DESC) rn
FROM yourTable
)
SELECT ID, MAX(M1) AS M1, MAX(M2) AS M2
FROM cte
WHERE rn = 1
GROUP BY ID;
The row number lets us restrict to only records for each ID having the most recent year date. Then, we aggregate to find the max values for M1 and M2.
In standard SQL, you can use lag(ignore nulls):
select id, coalesce(m1, prev_m1), coalesce(m2, prev_m2)
from (select t.*,
lag(m1 ignore nulls) over (partition by id order by date) as prev_m1,
lag(m2 ignore nulls) over (partition by id order by date) as prev_m2,
row_number() over (partition by id order by date desc) as seqnum
from t
) t
where seqnum = 1;
I'm having problem with getting only TOP 2 values for each group (groups are in column).
Example :
ID Group Value
1 A 30
2 A 150
3 A 40
4 A 70
5 B 0
6 B 100
7 B 90
I expect my output to be
ID Group Value
1 A 150
2 A 70
3 B 100
4 B 90
Simply, for each group I want just 2 rows with the highest Value
Most databases support the ANSI standard row_number() function. You would use it as:
select group, value
from (select t.*,
row_number() over (partition by group order by value desc) as seqnum
from t
) t
where seqnum <= 2;
To set the id you can use row_number() in the outer query:
select row_number() over (order by group, value) as id,
group, value
from (select t.*,
row_number() over (partition by group order by value desc) as seqnum
from t
) t
where seqnum <= 2;
However, changing the id seems suspicious.
You can use CTE with rank function ROW_NUMBER() .
Here is query to get your result.
;WITH cte AS
( SELECT Group, value,
ROW_NUMBER() OVER (PARTITION BY Group ORDER BY value DESC) AS rn
FROM test
)
SELECT Group, value FROM cte
WHERE rn <= 2
ORDER BY value
I've this data:
Id Date Value
'a' 2000 55
'a' 2001 3
'a' 2012 2
'a' 2014 5
'b' 1999 10
'b' 2014 110
'b' 2015 8
'c' 2011 4
'c' 2012 33
I want to filter out the first and the last value (when the table is sorted on the Date column), and only keep the other values. In case there are only two entries, nothing is returned. (Example for Id = 'c')
ID Date Value
'a' 2001 3
'a' 2012 2
'b' 2014 110
I tried to use order by (RANK() OVER (PARTITION BY [Id] ORDER BY Date ...)) in combination with this article (http://blog.sqlauthority.com/2008/03/02/sql-server-how-to-retrieve-top-and-bottom-rows-together-using-t-sql/) but I can't get it to work.
[UPDATE]
All the 3 answers seem fine. But I'm not a SQL expert, so my question is which one has the fastest performance if the table has around 800000 rows and there a no indexes on any column.
You can use row_number twice to determine the min and max dates and then filter accordingly:
with cte as (
select id, [date], value,
row_number() over (partition by id order by [date]) minrn,
row_number() over (partition by id order by [date] desc) maxrn
from data
)
select id, [date], value
from cte
where minrn != 1 and maxrn != 1
SQL Fiddle Demo
Here's another approach using min and max for this without needing to use a ranking function:
with cte as (
select id, min([date]) mindate, max([date]) maxdate
from data
group by id
)
select *
from data d
where not exists (
select 1
from cte c
where d.id = c.id and d.[date] in (c.mindate, c.maxdate))
More Fiddle
Here is a similar solution with row_number and count :
SELECT id,
dat,
value
FROM (SELECT *,
ROW_NUMBER()
OVER(
partition BY id
ORDER BY dat) rnk,
COUNT(*)
OVER (
partition BY id) cnt
FROM #table) t
WHERE rnk NOT IN( 1, cnt )
You can do this with EXISTS:
SELECT *
FROM Table1 a
WHERE EXISTS (SELECT 1
FROM Table1 b
WHERE a.ID = b.ID
AND b.Date < a.Date
)
AND EXISTS (SELECT 1
FROM Table1 b
WHERE a.ID = b.ID
AND b.Date > a.Date
)
Demo: SQL Fiddle