I have a table which looks like this:
ID
money_earned
days_since_start
1
1000
1
1
2000
2
1
3000
4
1
2000
5
2
1000
1
2
100
3
I want that rows, without a days_since_start (which means that the money_earned column was empty that day) - will include all the days PER ID, and fill the money_earned with last known value, so it to look like this:
ID
money_earned
days_since_start
1
1000
1
1
2000
2
1
2000
3
1
3000
4
1
2000
5
2
1000
1
2
1000
2
2
100
3
I have tried to look up for something like that, but I don't even know what function does that...
thank you!
You can try to use CTE RECURSIVE with OUTER JOIN and LAG window function to make it.
WITH RECURSIVE CTE
AS
(
SELECT ID,MIN(days_since_start) min_num,MAX(days_since_start) max_num
FROM T
GROUP BY ID
UNION ALL
SELECT ID,min_num+1,max_num
FROM CTE
WHERE min_num+1 <= max_num
)
SELECT c.ID,
CASE WHEN t1.ID IS NULL THEN LAG(money_earned) OVER(PARTITION BY c.ID ORDER BY c.min_num) ELSE money_earned END,
c.min_num days_since_start
FROM CTE c
LEFT JOIN T t1
ON c.min_num = t1.days_since_start
AND c.ID = t1.ID
ORDER BY c.ID
sqlfiddle
Related
I have a table that looks something like this:
customer_id
data
1
123
1
456
2
789
2
101
2
121
2
123
3
123
4
456
What I would like to do is perform a SELECT combined with a LIMIT X to get X number of records as well as any other records that have the same customer_id
Example query: SELECT customer_id, data FROM table ORDER BY customer_id LIMIT 3;
This query returns:
customer_id
data
1
123
1
456
2
789
I'd like a query that will look at the last customer_id value and return all remaining records that match beyond the LIMIT specified. Is it possible to do this in a single operation?
Desired output:
customer_id
data
1
123
1
456
2
789
2
101
2
121
2
123
In Postgres 13 can use with ties:
select t.*
from t
order by customer_id
fetch first 3 rows with ties;
In earlier versions you can use in:
select t.*
from t
where t.customer_id in (select t2.customer_id
from t t2
order by t2.customer_id
limit 3
);
You can use corelated subquery with count as follows:
Select t.*
From t
Where 3 >= (select count(distinct customer_id)
From t tt
where t.customer_id >= tt.customer_id)
My table
ID catone cattwo
100 2 1
100 3 1
200 1 2
expect result (count not sum)
ID totalcat1 totalcat2
100 2 2
200 1 1
My query
select COUNT(*) as totalcat1, catone
from Table1
group by cat1
left join
select COUNT(*) as totalcat2, cattwo
from Table1
group by cattwo
Try to have both count columns catone and cattwo
Not sure how to correct it. Thank you
A simple group-by should do it
select ID, COUNT(catone) as totalcat1, COUNT(cattwo) as totalcat2
from Table1
group by ID;
Note that this simply counts the number of values that are not NULL. If your original data was this...
ID catone cattwo
100 2 1
100 3 1
100 4 NULL
... then the result would be
ID totalcat1 totalcat2
100 3 2
If you want to count the distinct values - so totalcat2 would be 1 (as only 1 value exists in that column, although it's there twice) you could use
select ID, COUNT(DISTINCT catone) as totalcat1, COUNT(DISTINCT cattwo) as totalcat2
from Table1
group by ID;
which would return totalcat1 = 3 and totalcat2 = 1.
Here's a db<>fiddle with the two options.
Here's a second db<>fiddle on request of OP with ID 200.
Consider the following table Orders:
OrderID Name Amount
-----------------------
1 A 100
2 A 5
3 B 32
4 C 4000
5 D 701
6 E 32
7 F 200
8 G 100
9 H 12
10 I 17
11 J 100
12 J 100
13 J 11
14 A 5
I need to identify, for each unique 'Amount', if there are 2 or more users that have ordered that exact amount, and then list the details of those orders. So the desired output would be:
OrderID Name Amount
---------------------
1 A 100
8 G 100
11 J 100
12 J 100
3 B 32
6 E 32
please note that user A has ordered 2 x an order of 5 (order 2 and 14) but this shouldn't be in the output as it is within the same user. Only if another user would have made a order of 5, it should be in the output.
Can anyone help me out?
I would just use exists:
select o.*
from orders o
where exists (select 1
from orders o2
where o2.amount = o.amount and o2.name <> o.name
);
You can do :
select t.*
from table t
where exists (select 1 from table t1 where t1.amount = t.amount and t1.name <> t.name);
If you want only selected field then
SELECT Amount,name,
count(*) AS c
FROM TABLE
GROUP BY Amount, name
HAVING c > 1
ORDER BY c DESC
if you want full row
select * from table where Amount in (
select Amount, name from table
group by Amount, name having count(*) > 1)
I am trying to select the last change value per group.
I have a table
MMID column is incremental
MMID GID MID Value Bundle DateEntered
1 1 1 1 2 17/8/15 05:05:04
2 1 2 2 3 16/8/15 05:05:06
3 1 3 3 2 15/8/15 05:05:07
4 1 1 0 2 18/8/15 05:05:08
5 2 2 1 1 18/8/15 05:05:05
6 2 2 2 2 18/8/15 06:06:06
7 2 4 3 1 17/8/15 06:06:06
8 2 4 3 2 18/8/15 06:06:07
Here, I want the last change 'Value' in the last 24 hour(Having Date 18th August).
From the below query, I can get that. But even if the bundle value is changed, then I get that row.
But I want only rows when 'Value' is changed, or 'Value and Bundle' are changed. But not only when Bundle is changed
Desired output
MMID GID MID Value Bundle DateEntered
4 1 1 0 2 18/8/15 05:05:08
6 2 2 2 2 18/8/15 06:06:06
The query I tried is :
select yt1.*
from Table1 yt1
left outer join Table1 yt2
on (yt1.GID = yt2.GID and yt1.MID = yt2.MID
and yt1.MMID < yt2.MMID)
where yt2.MMID is null and yt2.GID is null and yt2.MID is null and yt1.DateEntered > '2015-08-18 00:00:00' ;
The output i get from here is:
MMID GID MID Value Bundle DateEntered
4 1 1 0 2 18/8/15 05:05:08
6 2 2 2 2 18/8/15 06:06:06
8 2 4 3 2 18/8/15 06:06:07
I should not be getting the last row here.
Can anyone tell me what should I change here.
Not really following the logic of your attempt, but here is how I would get the desired results:
WITH cte AS (
SELECT *,
ROW_NUMBER() OVER (PARTITION BY GID, MID ORDER BY MMID) AS rn
FROM Table
)
, cte2 AS (
SELECT t1.* FROM cte t1
INNER JOIN cte t2
ON t1.GID=t2.GID
AND t1.MID=t2.MID
AND t1.value<>t2.value
AND t1.rn=t2.rn+1
)
SELECT *
FROM cte2
WHERE MMID=(
SELECT TOP 1 MMID
FROM cte2 c2
WHERE cte2.GID=c2.GID
AND cte2.MID=c2.MID
ORDER BY MMID DESC
)
NB: If you don't want to include the rn column in the final results, use a column list instead of SELECT *.
id Name claim priority
1 yatin 70 5
6 yatin 1 10
2 hiren 30 3
3 pankaj 40 2
4 kavin 50 1
5 jigo 10 4
7 jigo 1 10
this is my table and i want to arrange this table as shown below
id Name claim priority AlternateFlag
1 yatin 70 5 0
6 yatin 1 10 0
2 hiren 30 3 1
3 pankaj 40 2 0
4 kavin 50 1 1
5 jigo 10 4 0
7 jigo 1 10 0
It is sorted as alternate group of same row.
I am Using sql server 2005. Alternate flag starts with '0'. In my example First record with name "yatin" so set AlternateFlag as '0'.
Now second record has a same name as "yatin" so alternate flag would be '0'
Now Third record with name "hiren" is single record, so assign '1' to it
In short i want identify alternate group with same name...
Hope you understand my problem
Thanks in advance
Try
SELECT t.*, f.AlternateFlag
FROM tbl t
JOIN (
SELECT [name],
AlternateFlag = ~CAST(ROW_NUMBER() OVER(ORDER BY MIN(ID)) % 2 AS BIT)
FROM tbl
GROUP BY name
) f ON f.name = t.name
demo
You could use probably an aggregate function COUNT() and then HAVING() and then UNION both Table, like:
SELECT id, A.Name, Claim, Priority, 0 as AlternateFlag
FROM YourTable
INNER JOIN (
SELECT Name, COUNT(*) as NameCount
FROM YourTable
GROUP BY Name
HAVING COUNT(*) > 1 ) A
ON YourTable.Name = A.Name
UNION ALL
SELECT id, B.Name, Claim, Priority, 1 as AlternateFlag
FROM YourTable
INNER JOIN (
SELECT Name, COUNT(*) as NameCount
FROM YourTable
GROUP BY Name
HAVING COUNT(*) = 1 ) B
ON YourTable.Name = B.Name
Now, this assumes that the Names are unique meaning the names like Yatin for example although has two counts is only associated to one person.
See my SqlFiddle Demo
You can use Row_Number() function with OVER that will give you enumeration, than use the reminder of integer division it by 2 - so you'll get 1s and 0s in your SELECT or in the view.