Minimum value without minus Value in SQL - ms-access

Column 1
Column 2
Column 3
5
-14
7
6
-20
7
7
-5
8
7
2
-1300
8
-9
8
9
-10
9
9
10
-1300
10
-15
9
10
15
-1300
With this query
select colum1, sum(column2), min(column3)
from Table
group by column1
We get
Column 1
Column 2
Column 3
5
-14
7
6
-20
7
7
-1
-1300
8
-9
8
9
0
-1300
10
0
-1300
How can I get value 8 except of -1300 in row 3, and value 9 except of -1300 in row 5, and value 9 except of -1300 in row 6 ?
I mean in column3, I need minimum value except minus value (like -1300 replace with other minimum value)
Column 1
Column 2
Column 3
5
-14
7
6
-20
7
7
-1
8
8
-9
8
9
0
9
10
0
9
I can't use operator because it will remove the row so column2 value will be change. It won't return the proper record


In standard SQL syntax, you would use case:
select column1, sum(column2),
min(case when column3 > 0 then colum3 end)
from Table
group by column1;
The equivalent in MS Access is:
select colum1, sum(colum2),
min(iif(column3 > 0, column3, null))
from Table
group by colum1

Related

How to add column in table grouped by value and max value in date column using sql query?

I have a table:
id date val
1 10.08.2022 10
1 12.08.2022 11
1 08.08.2022 15
1 16.08.2022 9
2 02.07.2022 2
2 01.07.2022 4
2 30.07.2022 7
I want to create two new columns last_v and max_v which are equal to last val for each id by date and maximum val per id. So desired output is:
id date val last_v max_v
1 10.08.2022 10 9 15
1 12.08.2022 11 9 15
1 08.08.2022 15 9 15
1 16.08.2022 9 9 15
2 02.07.2022 2 2 7
2 01.07.2022 4 2 7
2 30.06.2022 7 2 7
How could I do that?

You can use window functions!
select t.*,
first_value(val) over(partition by id order by dt desc) last_val,
max(val) over(partition by id) max_val
from mytable t
Demo on DB Fiddle:
id
dt
val
last_val
max_val
1
2022-08-08
15
9
15
1
2022-08-10
10
9
15
1
2022-08-12
11
9
15
1
2022-08-16
9
9
15
2
2022-06-30
7
2
7
2
2022-07-01
4
2
7
2
2022-07-02
2
2
7

Query with CASE statement in Group By I need to create an entry where no results occur in the WHEN

I have a query which groups up incoming payments into date ranges (1-7 days, 3-6 months etc.) and it largely works as I had hoped. However, I want to return a row which says 0 when no income is expected in the date range.
The group by looks like this:
group by
CASE WHEN timestampdiff(day,curdate(),data.duedate) between 0 and 7 then 1
WHEN timestampdiff(day,curdate(),data.duedate) between 8 and 14 then 2
WHEN timestampdiff(day,curdate(),data.duedate) between 15 and 30 then 3
WHEN timestampdiff(month,curdate(),data.duedate) between 1 and 2 then 4
WHEN timestampdiff(month,curdate(),data.duedate) between 2 and 3 then 5
WHEN timestampdiff(month,curdate(),data.duedate) between 3 and 6 then 6
WHEN timestampdiff(month,curdate(),data.duedate) between 6 and 12 then 7
WHEN timestampdiff(year,curdate(),data.duedate) between 1 and 2 then 8
WHEN timestampdiff(year,curdate(),data.duedate) between 2 and 3 then 9
WHEN timestampdiff(year,curdate(),data.duedate) between 3 and 4 then 10
WHEN timestampdiff(year,curdate(),data.duedate) between 5 and 6 then 11
WHEN timestampdiff(year,curdate(),data.duedate) >= 7 then 12
This works correctly in that it will give me the correct amounts, but I want to force the code to give me a 0. So I currently get this:
1 300000
5 150000
8 300000
What I actually want is this:
1 300000
2 0
3 0
4 0
5 150000
6 0
7 0
8 300000
etc.
This is the entire query - I've tried using an IFNULL() but had no success:
select
sum(data.principaldue+data.interestdue) as m
from
(select
la.id
,rep.duedate
,rep.PRINCIPALDUE
,rep.INTERESTDUE
from repayment rep
join loanaccount la on la.ENCODEDKEY = rep.PARENTACCOUNTKEY
join loanproduct lp on lp.ENCODEDKEY = la.PRODUCTTYPEKEY
group by
CASE WHEN timestampdiff(day,curdate(),data.duedate) between 0 and 7 then 1
WHEN timestampdiff(day,curdate(),data.duedate) between 8 and 14 then 2
WHEN timestampdiff(day,curdate(),data.duedate) between 15 and 30 then 3
WHEN timestampdiff(month,curdate(),data.duedate) between 1 and 2 then 4
WHEN timestampdiff(month,curdate(),data.duedate) between 2 and 3 then 5
WHEN timestampdiff(month,curdate(),data.duedate) between 3 and 6 then 6
WHEN timestampdiff(month,curdate(),data.duedate) between 6 and 12 then 7
WHEN timestampdiff(year,curdate(),data.duedate) between 1 and 2 then 8
WHEN timestampdiff(year,curdate(),data.duedate) between 2 and 3 then 9
WHEN timestampdiff(year,curdate(),data.duedate) between 3 and 4 then 10
WHEN timestampdiff(year,curdate(),data.duedate) between 5 and 6 then 11
WHEN timestampdiff(year,curdate(),data.duedate) >= 7 then 12
END
Order by
CASE WHEN timestampdiff(day,curdate(),data.duedate) between 0 and 7 then 1
WHEN timestampdiff(day,curdate(),data.duedate) between 8 and 14 then 2
WHEN timestampdiff(day,curdate(),data.duedate) between 15 and 30 then 3
WHEN timestampdiff(month,curdate(),data.duedate) between 1 and 2 then 4
WHEN timestampdiff(month,curdate(),data.duedate) between 2 and 3 then 5
WHEN timestampdiff(month,curdate(),data.duedate) between 3 and 6 then 6
WHEN timestampdiff(month,curdate(),data.duedate) between 6 and 12 then 7
WHEN timestampdiff(year,curdate(),data.duedate) between 1 and 2 then 8
WHEN timestampdiff(year,curdate(),data.duedate) between 2 and 3 then 9
WHEN timestampdiff(year,curdate(),data.duedate) between 3 and 4 then 10
WHEN timestampdiff(year,curdate(),data.duedate) between 5 and 6 then 11
WHEN timestampdiff(year,curdate(),data.duedate) >= 7 then 12
END

This is not a complete answer, but would be too big for comments;
A temporary table with numbers could be useful:
MySql temporary tables:
CREATE TEMPORARY TABLE TempTable (num int);
INSERT INTO TmpTable VALUES(1,2,3,4,5,6,7,8,9,10,11,12 ...);
Then you could right join on this table to make sure missing values are included.
Lets say you have this:
results(num, val):
1 300000
5 150000
8 300000
This should result in your desired output:
SELECT numbers.num, COALESCE(results.val, 0) as val
FROM results RIGHT JOIN TempTable numbers on results.num = numbers.num
WHERE numbers.num <= 12 --or other max number
1 300000
2 0
...
5 150000
...
Hope this helps.
Edit:
If you don't have permission to create temporary tables, look for a workaround to select consecutive integers, for example:
SELECT #row := #row + 1 as row, t.*
FROM some_table t, (SELECT #row := 0) r
Where some_table is any table with enough rows.
Probably use a top N on that.
Another dirty workaround, might be good enough if you don't need many numbers:
SELECT 1 num
UNION
SELECT 2 num
UNION
...
Edit:
Slightly tidier workaround:
SELECT * FROM (VALUES (1), (2), (3), ... ) x(i)

Finding continuous data from a table for months

I have a table which has month data in INT April=4, May=5 and so on. I want those records which have continuous data. My table is as follows. So if a record has discontinuous data, it should not be returned.
Continuous data means those records which having continuous four month. If records are there for 4,5,6,7 or 5,6,7,8 or 6,7,8,9 months then that record should come in result of a ID. If there records for a ID has 4,5,8,9 in month field this is discontinue data for me.
Initial query:
select ID, month from table1 where month in (4,5,6,7,8,9) group by month;
Initial Data:
PK ID month value
1 1 4 400
2 1 5 200
3 1 6 300
4 1 7 400
5 2 5 400
6 2 6 200
7 2 7 100
8 2 8 400
9 3 4 200
10 4 5 800
11 5 6 800
12 5 7 100
13 5 8 700
14 5 9 900
15 6 4 100
16 6 5 200
17 6 8 500
18 6 9 600
Result:
PK ID month value
1 1 4 400
2 1 5 200
3 1 6 300
4 1 7 400
5 2 5 400
6 2 6 200
7 2 7 100
8 2 8 400
11 5 6 800
12 5 7 100
13 5 8 700
14 5 9 900
My database is MySQL.
I have months from April(4) to Sept(9)

Try the below query this will work I hope!
SELECT ID, Month
FROM table1 WHERE month in (4,5,6,7,8,9)
GROUP BY ID, Month
HAVING count(ID)>3
kindly let me know this is working or not

Puzzle: find number in a given row and column in 2-dimensional array where no number may occur twice

We have a two-dimensional array with the number 0 in the upper left corner. The rest of the array is then filled with numbers so that each index contains the smallest positive integer possible that already exists neither on the same row or column.
Example:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
1 0 3 2 5 4 7 6 9 8 11 10 13 12 15 14 17 16
2 3 0 1 6 7 4 5 10 11 8 9 14 15 12 13 18 19
3 2 1 0 7 6 5 4 11 10 9 8 15 14 13 12 19 18
4 5 6 7 0 1 2 3 12 13 14 15 8 9 10 11 20 21
5 4 7 6 1 0 3 2 13 12 15 14 9 8 11 10 21 20
6 7 4 5 2 3 0 1 14 15 12 13 10 11 8 9 22 23
7 6 5 4 3 2 1 0 15 14 13 12 11 10 9 8 23 22
8 9 10 11 12 13 14 15 0 1 2 3 4 5 6 7 24 25
9 8 11 10 13 12 15 14 1 0 3 2 5 4 7 6 25 24
10 11 8 9 14 15 12 13 2 3 0 1 6 7 4 5 26 27
11 10 9 8 15 14 13 12 3 2 1 0 7 6 5 4 27 26
12 13 14 15 8 9 10 11 4 5 6 7 0 1 2 3 28 29
13 12 15 14 9 8 11 10 5 4 7 6 1 0 3 2 29 28
14 15 12 13 10 11 8 9 6 7 4 5 2 3 0 1 30 31
15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 31 30
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 0 1
17 16 19 18 21 20 23 22 25 24 27 26 29 28 31 30 1 0
Given the row and the column in such array, I need to be able to find the number in the specified index in less than one second on a relatively new desktop PC (for row and column less than a million). My brute-force attempts so far have been so futile that it's clearly not the way I want to go with this. Presumably there must be a way to find out the number in question, in linear time (?), that doesn't require computing all the preceding numbers in the array.

Observation shows that the operator is the bitwise XOR (represent each operand as a binary number, XOR together the corresponding bits, read as binary).
Now on to prove it is the XOR:
Since the XOR with one argument fixed is a bijection on the other argument, the "that exists neither on the same row or column" is satisfied.
Now it just suffices to prove the "smallest" part, namely that any lower value already occurs if we reduce either operand:
foreach A >= 0, B >= 0, F >= 0:
(A xor B > F) => (exists D: D xor B = F) or (exists E: A xor E = F)
or equivalently
foreach 0 <= A, 0 <= B, 0 <= F < (A XOR B)
(exists D: D xor B = F) or (exists E: A xor E = F)
Note that we are no longer concerned about our operator, we're proving the minimality of XOR.
Define C = A xor B
if A = 0, B = 0, then minimality is satisfied.
Now, if A and B have the same magnitude (the same bit length), then clearing the top bit of both will not change C. Clearing the top bit is a translation towards the origin in the matrix, so if a smaller value exists above or to the left after translation, it is at the same relative position before the translation.
A and B must have a different magnitude to be a counter-example. XOR (as well as the operator under consideration) are symmetric, so assume A > B.
If F is of greater magnitude than A, then it's not smaller, and thus it's not a counter-example.
If F has the same magnitude as A, then clear the highest bit in A and in F. This is a translation in the table. It changes the values, but not their ordering, so if a smaller value exists above or to the left after translation, it is at the same relative position before the translation.
If F has a smaller magnitude than A, then, by the pigeonhole principle and the properties of XOR, there exists a D with a smaller magnitude than A such that D xor B = F.
summary: The proof that XOR satisfies the conditions imposed onto the solution follows from the symmetries of XOR, its magnitude-preserving properties and its bijection properties. We can find each smaller element than A xor B by reducing A, B and the challenge until they're all zero or of different magnitude (at which point we apply the pigeonhole principle to prove the challenge can be countered without actually countering it).

Any Advice with Mysql Count

i have a t_class table in mySql,
in this table there are 3 columns, No, CLASS and POINT.
and there are approximately 5000 records in this table. i want the count of classes in this table.
No CLASS POINT
1 9 100
2 10 70
3 11 80
4 9 90
5 10 50
6 M 60
7 M 70
8 9 40
9 10 90
10 11 90
11 M 80
12 M 75
13 11 40
14 10 100
15 9 60
As you see there 4 types of classes - 9, 10, 11 and M.
But there is one problem. When it calculates the count of classes it must
summarize 11-th and M th classes. For example
CLASS COUNT
9 4
10 4
11 7
Thanks.

SELECT CLASS, COUNT(*) AS CNT
FROM table
GROUP BY CASE WHEN CLASS='M' THEN '11' ELSE CLASS END