I've tried more or less all combinations of count and distinct (except the correct one :) ) in order to get the example below.
Input: table t1
NAME | FOOD
Mary | Apple
Mary | Banana
Mary | Apple
Mary | Strawberry
John | Cherries
Expected output:
NAME | FOOD
Mary | 3
John | 1
N.B. Mary has Apple in two rows but she has 3 as we have 3 different values in the column.
I only managed to get 4 in FOOD Column for her, but I need 3 :(
select a.name as NAME, a.count(name) as Food
from
(SELECT distinct NAME,Food from table)a
Start with a query which gives you unique combinations of NAME and FOOD:
SELECT DISTINCT t1.NAME, t1.FOOD
FROM t1
Then you can use that as a subquery in another where you can GROUP BY and Count:
SELECT sub.NAME, Count(*) AS [FOOD]
FROM
(
SELECT DISTINCT t1.NAME, t1.FOOD
FROM t1
) AS sub
GROUP BY sub.NAME;
select a.name, sum(a.FoodCount) from(
select distinct name,COUNT(food) as FoodCount from #t1 group by name, food ) as a group by a.name order by 2 desc
Related
Consider this table:
id name department email
1 Alex IT blah#gmail.com
1 Alex IT blah#gmail.com
2 Jay HR jay#gmail.com
2 Jay Marketing zou#gmail.com
If I group byid,name and count I get:
id name count(*)
1 Alex 2
2 Jay 2
With this query:
select id,name,count(*) from tb group by id,name;
However I would like to count only records that diverge from department,email, so as to have:
id name count(*)
1 Alex 0
2 Jay 1
This time the count for the first group 1,Alex is 0 because department,email have the same values (duplicated) , on the other hand 2,Jay is one because department,email has one different value.
If you meant "two different values" for "Jay", you can use distinct:
select id,name,count(*) from (SELECT distinct * FROM tb) group by id,name;
You can use count(*) - 1 to get similar results in your question.
Currently I have a table this :
Roll no. Names
------------------
1 Sam
1 Sam
2 Sasha
2 Sasha
3 Joe
4 Jack
5 Jack
5 Julie
I want to write a query in which I get count of the combination in another column
Required output
Combination distinct count
-----------------------------
2-Sasha 1
5-Jack 1
5-Julie 1
Basically, you could group by these columns and use a count function:
SELECT rollno, name, COUNT(*)
FROM mytable
GROUP BY rollno, name
You could also concat the two columns:
SELECT CONCAT(rollno, '-', name), COUNT(*)
FROM mytable
GROUP BY CONCAT(rollno, '-', name)
I am hoping someone can help me with my query.
I have a table with the columns, 'Date', 'ID_Num and 'Name'. What I want to do is add a column at the end to show the total amount of times each ID_Num is within the data but based on the date. So although 'ID_Num' 1001 shows 4 times in total, it is twice on the 20/04/2018 and once on both the 21/04/2018 and 22/04/2018.
EDIT: I should have stipulated that I will be pulling several other columns with information, which I cant use a group by on everything.
Date ID_Num Name Count
20/04/2018 1001 John 2
20/04/2018 1001 John 2
20/04/2018 1002 Paul 2
20/04/2018 1002 Paul 2
20/04/2018 1003 David 2
20/04/2018 1003 David 2
20/04/2018 1004 Stephen 1
21/04/2018 1001 John 1
21/04/2018 1002 Paul 3
21/04/2018 1002 Paul 3
21/04/2018 1002 Paul 3
21/04/2018 1004 Stephen 1
22/04/2018 1001 John 1
22/04/2018 1002 Paul 1
22/04/2018 1003 David 1
22/04/2018 1004 Stephen 1
Thanks
Unless I'm missing something here, a simple group by and count should do it:
SELECT Date, ID_Num, Name, Count(*)
FROM TableName
GROUP BY Date, ID_Num, Name
(That is, assuming there can only be one Name for each ID_Num)
Update
Assuming your rdbms supports it, you can use count with an over clause:
SELECT Date, ID_Num, Name, Count(*) OVER(PARTITION BY Date, Id_Num)
FROM TableName
If not, you can use a sub query:
SELECT Date,
ID_Num,
Name,
(SELECT Count(*)
FROM TableName As t1
WHERE t1.Date = t0.Date
AND t1.ID_NUM = t0.ID_NUM)
FROM TableName As t0
Try this:
SELECT
Date,
Id_num,
count(*) count
FROM
tabel_name
GROUP BY
Date,
Id_num
If you want name as well:
SELECT
Date,
Id_num,
Name
count(*) count
FROM
tabel_name
GROUP BY
Date,
Id_num,
Name
You can use a normal select query and then add a sub query to do a group and show the total. Simple example below
SELECT Date, ID_Num, Name,
(SELECT Count(ID_Num) FROM TableName AS CHILD WHERE CHILD.Id_Num = Parent.Id_Num) AS Total
FROM TableName AS Parent
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
Here is an (simplified) example of DB I have (sorry for the ulgy format, I don't know how to write tables):
Name | Num
John | 1
John | 3
John | 4
Dany | 2
Andy | 5
Andy | 5
I want to count how many people have more at least two different Numbers.
For instance, here, only john, because he has 1, 3 and 4.
Not Andy because he has twice 2 and no other one.
And obviously not Dany because he has only one entry.
Thank you very much.
Try this.
select count(name) from table group by name having count(distinct num)>1
Try this:
SELECT A.Name, COUNT(DISTINCT A.Num) cnt
FROM tableA
GROUP BY A.Name
HAVING cnt >= 2;
select count(*)
from (
select Name from Temp group by Name having count(distinct num) > 1
) as a
Try this:
select name from `table` group by name,num having count(num)>1