I am pretty new to SQL and I thought I was comfortable using it after a while but it still is tough. I am trying to increment ids. I know I could use auto-increment but in this case there are id has relationship with several categories so it has to start with different numbers so I can't do it.
The table looks something like this:
id category
----------------
1000 1
1000 1
...
2000 2
2000 2
...
And I want to make it:
id category
------------------
1000 1
1001 1
1002 1
...
2000 2
2001 2
...
I tried:
UPDATE T1
SET id = CASE
WHEN EXISTS (SELECT id FROM STYLE WHERE T1.id = id)
THEN (SELECT MAX(CAST(id AS INT)) + 1
FROM STYLE
WHERE category = T1.category)
END
FROM STYLE T1
WHERE idStyle = idStyle
But it just added 1 to all rows. How could I go 1 by 1 so it could actually get the incremented max id? Thank you.
In the absense of real sample data, this is a pseudo-sql, however, something like...
UPDATE YT
----SELECT NULL as Ihave no context of other fields in your table
SET id = id + ROW_NUMBER() OVER (PARTITION BY category ORDER BY (SELECT NULL)) - 1
FROM YourTable YT;
You can use row_number() function instead :
select *,
concat(cid, row_number() over (partition by id order by category)-1) as NewId
from style s;
Related
I have table in which I inserted all the liked places records.
Like:
I have table PlaceLikes;
Id placeId likedByUserID
1 ABC 1
2 DEF 1
3 ABC 2
4 FFF 2
Result: User 1 want to get all placeID that matches with itself.
Id placeId likedByUserID
3 ABC 2
Here User 2 with ABC placeId is similar with Requestor User ID 1.
So, How I can filter the data like this
You can use exists:
select t.*
from mytable t
where
t.likedByUserID <> 1
and exists (
select 1 from mytable t1 where t1.place_id = t.place_id and t1.likedByUserID = 1
)
I think this piece of query will be able to solve your problem statement as understood by me.
DECLARE #sample_name VARCHAR (10);
SET #sample_name = 'placeId to be searched'
SELECT Id, placeId, likedByUserID FROM demo_table dt
WHERE dt.placeId = #sample_name
Do let me know if this was helpful.
I have the task to find out if all columns in a SQL Server table have exact the same value. The table content is created by a stored procedure and can vary in the number of columns. The first column is an ID, the second and the following columns must be compared if the all columns have exact the same value.
At the moment I do not have a clue how to achieve this.
The best solution would be to display only the rows, which have different values in one or multiple columns except the first column with ID.
Thank you so much for your help!!
--> Edit: The table looks this:
ID Instance1 Instance2 Instance3 Instance4 Instance5
=====================================================
A 1 1 1 1 1
B 1 1 0 1 1
C 55 55 55 55 55
D Driver Driver Driver Co-driver Driver
E 90 0 90 0 50
F On On On On On
The result should look like this, only the rows with one or multiple different column values should be display.
ID Instance1 Instance2 Instance3 Instance4 Instance5
=====================================================
B 1 1 0 1 1
D Driver Driver Driver Co-driver Driver
E 90 0 90 0 50
My table has more than 1000 rows and 40 columns
you can achieve this by using row_number()
Try the following code
With c as(
Select id
,field_1
,field_2
,field_3
,field_n
,row_number() over(partition by field_1,field_2,field_3,field_n order by id asc) as rn
From Table
)
Select *
From c
Where rn = 1
row_number with partition is going to show you if the field is repeated by assigning a number to a row based on field_1,field_2,field_3,field_n, for example if you have 2 rows with same field values the inner query is going to show you
rn field_1 field_2 field_3 field_n id
1 x y z a 5
2 x y z a 9
After that on the outer part of the query pick rn = 1 and you are going to obtain a query without repetitions based on fields.
Also if you want to delete repeated numbers from your table you can apply
With c as(
Select id
,field_1
,field_2
,field_3
,field_n
,row_number() over(partition by field_1,field_2,field_3,field_n order by id asc) as rn
From Table
)
delete
From c
Where rn > 1
The best solution would be to display only the rows, which have different values in one or multiple columns except the first column with ID.
You may be looking for a the following simple query, whose WHERE clause filters out rows where all fields have the same value (I assumed 5 fields - id not included).
SELECT *
FROM mytable t
WHERE NOT (
field1 = field2
AND field1 = field3
AND field1 = field4
AND field1 = field5
);
I have a table
id number name update_date
1 123 asd 08.05.18
2 412 ddd 08.05.18
3 123 dsa 14.05.18
4 125 dsa 05.05.18
Whole table consist from that rows like that. I need to select row 1 and 3 because I need different update_dates but same number. How to do that? I need to see the changes from specific Number between 2 update dates 08.05.18 and 14.05.18. I have more update dates in my table.
I tried:
SELECT *
FROM legal_entity_history a
JOIN legal_entity_history b ON a.BIN = b.BIN
WHERE ( a.update_date <> b.update_date AND
a.update_date = "08.05.18" AND
b.update_date = "14.05.18" )
A relatively simple method is:
select leh.*
from legal_entity_history leh
where exists (select 1
from legal_entity_history leh2
where leh2.number = leh.number and leh2.update_date <> leh.update_date
);
For performance, you want an index on legal_entity_history(number, update_date).
TRY THIS: Assuming that same number may not appear more than once under same update_date, so, you can achieve that using GROUP BY with HAVING as below
SELECT t.*
FROM test t
INNER JOIN (SELECT number
FROM test
GROUP BY number
HAVING COUNT(DISTINCT update_date) > 1) t1 ON t1.number = t.number
OUTPUT:
id number name update_date
1 123 asd 08.05.18
3 123 dsa 14.05.18
I have a table like so
Id | Type | Value
--------------------
0 | Big | 2
1 | Big | 3
2 | Small | 3
3 | Small | 3
I would like to get a table like this
Type | Last Value
--------------------
Small | 3
Big | 3
How can I do this. I understand there is an SQL Server method called LAST_VALUE(...) OVER .(..) but I can't get this to work with GROUP BY.
I've also tried using SELECT MAX(ID) & SELECT TOP 1.. but this seems a bit inefficient since there would be a subquery for each value. The queries take too long when the table has a few million rows in it.
Is there a way to quickly get the last value for these, perhaps using LAST_VALUE?
You can do it using rownumber:
select
type,
value
from
(
select
type,
value,
rownumber() over (partition by type order by id desc) as RN
) TMP
where RN = 1
Can't test this now since SQL Fiddle doesn't seem to work, but hopefully that's ok.
The most efficient method might be not exists, which uses an anti-join for the underlying operator:
select type, value
from likeso l
where not exists (select 1 from likeso l2 where l2.type = l.type and l2.id > l.id)
For performance, you want an index on likeso(type, id).
I really wonder if there is more efficent solution but, I use following query on such needs;
Select Id, Type, Value
From ( Select *, Max (Id) Over (Partition By Type) As LastId
From #Table) T
Where Id = LastId
Code Description Whatever
---------------------------------
1 stuff blah
1 something meh
2 yah bong
2 never hammer time
How do I get a results set from this with each Code only present once? (I don't overly care which record for that code it is).
So I want....
1 stuff blah
2 yah bong
SELECT *
FROM (
SELECT * , row_number() over(partition by code order by Description) as id
from yourTable
) temp
WHERE id = 1
I think this is sql server only
You first need to pick a column which determines what counts as 'the first result'. In my example I chose Description:
SELECT * FROM YourTable first
WHERE
(SELECT COUNT(*) FROM YourTable previous
WHERE previous.Code=first.Code AND previous.Description < first.Description) = 0