There is mysql Ver 8.0.18 value_table as:
value count
1 3
11 1
12 2
22 5
31 1
34 3
35 1
40 3
46 7
What is query to get a total count for each dozen (1-10 - first dozen,11-20 - second , etc..)
as:
1 3
2 3
3 5
4 8
5 7
Query should be flexible, so when some records added to value_table , for example
51 2
62 3
so, it is not necessary to change a query by adding new range (51-60 - 6-th dozen, etc.)
I think you just want division and aggregation:
select min(value), sum(count)
from t
group by floor(value / 10);
To be honest, I'm not sure if the first column should be min(value) or floor(value / 10) + 1.
Related
i have a table :
a b c
1 10 1001
7 6 54
56 2000 31
1200 5 9
4 10 20
2 65 20
how can i select rows with column's value of this row smaller than 1000. i want to get this
a b c
7 6 54
4 10 20
2 65 20
mysql query still get all value :
SELECT a,b,c FROM test
where a <'1000' or b<'1000' or c<'1000'
It sounds like you would like to pull a row where there is NO column greater than 1000 in that row, if that is correct then you need to us AND instead of OR.
SELECT a,b,c FROM test
where a <'1000' AND b<'1000' AND c<'1000'
Hope that helps!
In mysql, I need a query that returns the quantity of repeated values in the field "Info" of my table "Log".
Table Log:
ID_Log User Info
1 1 3
2 1 3
3 1 3
4 1 5
5 1 6
6 1 6
7 1 7
8 1 8
9 1 8
The query should return "4" (Info 3 appears three times, Info 6 appears two times, Info 8 appears two times).
Any suggestions?
You can get the number of values that have already appeared by using a simple subtraction. Subtract the number of distinct values from the total number of rows:
select count(*) - count(distinct info)
from log;
The difference is the number that "repeat".
This should work. Group the values of info together and only keep the results where the number of occurrences minus 1 is greater than 0. Then sum the numbers of occurrences.
select sum(repeats)
from (SELECT Info, count(*) - 1 AS repeats
FROM Log
GROUP BY Info
HAVING repeats > 0)
Here is a problem. I need to find working days between two dates (without weekends). I currently found and was able to successfully use this approach:
SELECT 5 * (DATEDIFF(DateClosed, DateOpened) DIV 7) + MID('0123455501234445012333450122234501101234000123450',
7 * WEEKDAY(DateOpened) + WEEKDAY(DateClosed) + 1, 1) AS TotalResolutionTimeBusinessDays
FROM table1
This is a complex calculation based on a matrix:
| M T W T F S S
-|--------------
M| 0 1 2 3 4 5 5
T| 5 0 1 2 3 4 4
W| 4 5 0 1 2 3 3
T| 3 4 5 0 1 2 2
F| 2 3 4 5 0 1 1
S| 0 1 2 3 4 0 0
S| 0 1 2 3 4 5 0
In the matrix, the intersection of any given x and y value pair (WEEKDAY(#S) and WEEKDAY(#E) yields the difference in work days between the two values. The 49 values in the table are concatenated into the following string: 0123455501234445012333450122234501101234000123450.
This works fine for me but now I need to present difference between dates in another format. E.g.: Let's say we have two dates: StartDate = '2013-06-28 01:27:35' and EndDate = '2013-07-08 16:47:21'. If we use method described above we get 7 working days which is correct. But I need to count all the difference between dates(including hours and minutes) so it could look like
SELECT TIME_TO_SEC(TIMEDIFF('2013-07-08 16:47:21','2013-06-28 01:27:35')) / 3600 / 24
which without weekends should be value like 7.64 days.
Any suggestions how to write a calculation based on that format? Any help would be much appreciated.
I have one employee table which contains:
emp id Sum
------ ---
1 7
2 6
I want a SQL query for getting the quotient and remainder when dividing the Sum with 8.
Use integer division and mod operators to get the quotient and remainder:
SELECT
emp_id,
sum,
sum / 8 AS Result,
sum div 8 AS Quotient,
sum mod 8 AS Remainder
FROM employee
emp_id sum Result Quotient Remainder
1 7 0.8750 0 7
2 6 0.7500 0 6
3 9 1.1250 1 1
4 10 1.2500 1 2
5 11 1.3750 1 3
6 12 1.5000 1 4
7 13 1.6250 1 5
8 14 1.7500 1 6
9 15 1.8750 1 7
10 16 2.0000 2 0
What will be the return type of your qoutient? If you don't care if its a floating point or an integer(whole number). You can try this.
SELECT
(sum / 8) AS qoutient,
(sum % 8) AS reminder
FROM employee
You can use the mysql function DIV to get the qoutient
(http://dev.mysql.com/doc/refman/5.0/en/arithmetic-functions.html#operator_div) :
SELECT 14 DIV 3
will return 4
It should do the trick
As for the remainder, others have replied
you can use the % operator to get the remainder. Here's an example.
SELECT Round(17 / 4) -- quotient without decimal
SELECT 17 % 4 -- remainder
For my PL/SQL function I usually use:
SELECT trunc(sum/8) --- Quotient
SELECT mod(sum/8) -- Remainder
use floor function, select floor(column_name/divisor) from dual
I have the table below
relID value charge
1 2 5
1 8 2
2 1 10
2 4 6
2 9 2
For the above table i need for a given value ex 10 to find what to charge for each relID
In the above for value<10 i need to get charge=5 for relID=1 and charge=2 for relID=2
I am trying to use 1 sql command to get it and i am kind of lost
Can anyony help
Thanks
Your question isn't very clear but I think this will work for you
SELECT t.relID,
(
SELECT charge
FROM table
WHERE relID = t.relID
AND value < 10
ORDER BY value
LIMIT 1
) AS charge
FROM table AS t
Let me rephrase.
Here is the table
relID value charge
1 2 5
1 8 2
2 1 10
2 4 6
2 9 2
Explain the table :
Lets say that the value and charge are money.
If the user has value 2 then i must charge with 5 using relID 1
If the user has value 8 then i must charge with 2 using relID 1
same for the relID 2
So when a user come with a value 10 i must find what to charge.So for the given value 10 i must find in the table all records with value<10.
In the example the values for value <10 are
For relID=1 are (2,8)
For relID=2 are (1,4,9)
Now for each relID i need to get the max value.
For relID=1 max value is 8 so charge is 2
For relID=2 max value is 9 so charge is 2
I plain english there are value rate
0-2 charge 5
2-8 charge 2
and ...
i hope to be clear now
select *
from Table t
where value =
(select max(value)
from Table
where value <= 10
and relId = t.relId)