I could get the minimum percentage of two values, but I need only the name, and ID in the select.
ID NAME CITY ONE TWO
--------------------------------------------------
2 Morales Los Angeles 40 10
1 John New York 60 20
4 Mary San Diego 10 10
I need to get the min value of one/two, and to only appear this as a result:
ID NAME
---------
4 Mary
Select ID, NAME
from MYTABLE
where least(ONE,TWO) = (select min(least(ONE,TWO)) from MYTABLE);
If you don't want Morales, then you can do this :
Select ID, NAME
from MYTABLE
where id =
(select id from
(select id from MYTABLE order by least(ONE,TWO), ONE*TWO)
where rownum <= 1);
Related
The simple SELECT query would return the data as below:
Select ID, User, Country, TimeLogged from Data
ID User Country TimeLogged
1 Samantha SCO 10
1 John UK 5
1 Andrew NZL 15
2 John UK 20
3 Mark UK 10
3 Mark UK 20
3 Steven UK 10
3 Andrew NZL 15
3 Sharon IRL 5
4 Andrew NZL 25
4 Michael AUS 5
5 Jessica USA 30
I would like to return a sum of time logged for each user grouped by ID
But for only ID numbers where both of these values Country = UK and User = Andrew are included within their rows.
So the output in the above example would be
ID User Country TimeLogged
1 John UK 5
1 Andrew NZL 15
3 Mark UK 30
3 Steven UK 10
3 Andrew NZL 15
First you need to identify which IDs you're going to be returning
SELECT ID FROM MyTable WHERE Country='UK'
INTERSECT
SELECT ID FROM MyTable WHERE [User]='Andrew';
and based on that, you can then filter to aggregate the expected rows.
SELECT ID,
[User],
Country,
SUM(Timelogged) as Timelogged
FROM mytable
WHERE (Country='UK' OR [User]='Andrew')
AND ID IN( SELECT ID FROM MyTable WHERE Country='UK'
INTERSECT
SELECT ID FROM MyTable WHERE [User]='Andrew')
GROUP BY ID, [User], country;
So, you have described what you need to write almost perfectly but not quite. Your result table indicates that you want Country = UK OR User = Andrew, rather than AND
You need to select and group by, then include a WHERE:-
Select ID, User, Country, SUM(Timelogged) as Timelogged from mytable
WHERE Country='UK' OR User='Andrew'
Group by ID, user, country
I have one database and time to time i change some part of query as per requirement.
i want to keep record of results of both before and after result of these queries in one table and want to show queries which generate difference.
For Example,
Consider following table
emp_id country salary
---------------------
1 usa 1000
2 uk 2500
3 uk 1200
4 usa 3500
5 usa 4000
6 uk 1100
Now, my before query is :
Before Query:
select count(emp_id) as count,country from table where salary>2000 group by country;
Before Result:
count country
2 usa
1 uk
After Query:
select count(emp_id) as count,country from table where salary<2000 group by country;
After Query Result:
count country
2 uk
1 usa
My Final Result or Table I want is:
column 1 | column 2 | column 3 | column 4 |
2 usa 2 uk
1 uk 1 usa
...... but if query results are same than it shouldn't show in this table.
Thanks in advance.
I believe that you can use the same approach as here.
select t1.*, t2.* -- if you need specific columns without rn than you have to list them here
from
(
select t.*, row_number() over (order by count) rn
from
(
-- query #1
select count(emp_id) as count,country from table where salary>2000 group by country;
) t
) t1
full join
(
select t.*, row_number() over (order by count) rn
from
(
-- query #2
select count(emp_id) as count,country from table where salary<2000 group by country;
) t
) t2 on t1.rn = t2.rn
Suppose following table:
Name Age Occupation
Alex 20 Student
Alex 20 Seller
Alex 20 Minister
Liza 19 Student
Liza 20 Volunteer
Liza 21 HR partner
I want to find names which have only (and only) 20 in age column. So from this table I want to get all "Alex" rows and no "Liza" rows at all.
Thanks!
You need to use Group By and Having clause. Try this way
select Name
from table
group by Name
having count(case when Age = 20 then 1 end) = count(*)
count(case when Age = 20 then 1 end) counts only when age = 20 if it is equal to total count then the name has only 20 as age.
Just one another way:
select Name
from table
group by Name
having min(Age) = 20 and max(Age) = 20
One way is using NOT IN():
SELECT Name, Age, Occupation
FROM YourTable
WHERE Age = 20
AND Name NOT IN (SELECT Name FROM YourTable WHERE Age <> 20)
I'd like to order pairs (or group of 3,4 etc.) of rows given the SUM of a certain value.
The rows are consecutive based on the concatenation of Name+Surname+Age
To better understand given the following table:
ID Name Surname Age Salary
------------------------------
1 John Smith 30 2
2 John Smith 30 10
3 Rick James 22 300
4 Rick James 22 1000
5 Rick James 22 5
6 Mike Brown 50 200
7 Mike Brown 50 20
I'd like to have a final table that should be ordered DESC by the sum of Salary of each Name+Surname+Age and keeping the rows with same Name+Surname+Age next to each others despite the ID column is different. This would be the expected result:
ID Name Surname Age Salary
------------------------------
3 Rick James 22 300
4 Rick James 22 1000
5 Rick James 22 5
6 Mike Brown 50 200
7 Mike Brown 50 20
1 John Smith 30 2
2 John Smith 30 10
As you can see the rows with Name+Surname+Age = "Rick Jams 22" are on the top since their total sum would be 1305, followed by "Mike Brown 50" (sum = 220) and "John Smith 30" (sum = 12).
Additionally, the number of rows has to be the same in the resulting table.
How can I do that using Oracle SQL?
Thanks for any help
SELECT t.*,
COALESCE(SUM(salary) OVER (PARTITION BY name, surname, age), 0) ss
FROM mytable t
ORDER BY
ss DESC
Try this:
SELECT ID, Name, Surname, Age, Salary
FROM (
SELECT ID, Name, Surname, Age, Salary,
SUM(Salary) OVER (PARTITION BY Name, Surname, Age) AS sum_of_sal
FROM mytable) t
ORDER BY sum_of_sal DESC, ID
The query uses the window version of SUM in order to calculate the sum of salaries per Name, Surname, Age partition. We can use this field in an outer query to do the sorting.
or try this
SELECT ID, Name, Surname, Age, Salary
FROM mytable
ORDER BY SUM(Salary) OVER (PARTITION BY Name, Surname, Age) DESC, ID
I have a table sample table as follows:
ID | City
--------------
1 | New York
2 | San Francisco
3 | New York
4 | Los Angeles
5 | Atlanta
I would like to select the distinct City AND the TOP ID for each. E.g., conceptually I would like to do the following
SELECT TOP 1 ID, DISTINCT City
FROM Cities
Should give me:
ID | City
--------------
1 | New York
2 | San Francisco
4 | Los Angeles
5 | Atlanta
Because New York appears twice, it's taken the first ID 1 in this instance.
But I get the error:
Column 'Cities.ID' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.
Try this way:
SELECT min(ID), City
FROM Cities
Group by City
MIN function is used for choose one of the ID from two New York cities.
You need to have your city in a GROUP BY
SELECT MIN(ID), City
FROM Cities
GROUP BY City
More general solution is to use row_number in order to get other details of table:
select * from
(select *, row_number() over(partition by City order by ID) as rn from Cities)
where rn = 1
But for this particular table just grouping will do the work:
select City, Min(ID) as ID
from Cities
group by City
If you have a complex scenario where Group By cannot use, You could use Row_Number() function with Common Table Expression.
;WITH CTE AS
(
SELECT ID, City, ROW_NUMBER() OVER (PARTITION BY City ORDER BY Id) rn
FROM YourTable
)
SELECT Id, City
FROM CTE
WHERE rn = 1