I have a table something like this
Id student_name City
4 abc Mumbai
6 xyz Delhi
4 lmn Kolkata
6 abc Mumbai
6 GHI Chennai
I am using dense_rank() function to dismiss the duplicate entry of ID in the table means if I am having the ID 4 twice it should give me only once in the output.
When I am using dense_rank function like:
select dense_rank() over (order by student_id desc ) as ID ,Id, student_name,city
from test
It is giving me the output something like this
ID ID student_name city
1 4 abc Mumbai
1 4 lmn kolkata
2 6 xyz Delhi
2 6 abc Mumbai
But I don't want duplicate how to remove using dense_rank() function
First, dense_rank() in the select does not do filtering. So, I don't know why the output would have four rows when the input has five.
Second, to keep only one row per id, then use row_number(), not dense_rank() . . . along with a subquery:
select t.*
from (select t.*,
row_number() over (partition by student_id order by student_id) as seqnum
from test t
) t
where seqnum = 1;
Related
I'm trying to match and align data, or resaid, count occurrences and then list for which values those occurrences occur.
Or, in a question: "How many times does each ID value occur, and for what names?"
For example, with this input
Name ID
-------------
jim 123
jim 234
jim 345
john 123
john 345
jane 234
jane 345
jan 45678
I want the output to be:
count ID name name name
------------------------------------
3 345 jim john jane
2 123 jim john
2 234 jim jane
1 45678 jan
Or similarly, the input could be (noticing that the ID values are not aligned),
jim john jane jan
----------------------------
123 345 234 45678
234 123 345
345
but that seems to complicate things.
As close as I am to the desired results is in SQL, as
for ID, count(ID)
from table
group by (ID)
order by count desc
which outputs
ID count
------------
345 3
123 2
234 2
45678 1
I'll appreciate help.
You seem to want a pivot. In SQL, you have to specify the number of columns in advance (unless you construct the query as a string).
But the idea is:
select ID, count(*) as cnt,
max(case when seqnum = 1 then name end) as name_1,
max(case when seqnum = 2 then name end) as name_2,
max(case when seqnum = 3 then name end) as name_3
from (select t.*,
row_number() over (partition by id order by id) as seqnum -- arbitrary ordering
from table t
) t
group by ID
order by count desc;
If you have an unknown number of columns, you can aggregate the values into an array:
select ID, count(*) as cnt,
array_agg(name order by name) as names
from table t
group by ID
order by count desc
the query would look similar to this if that's what you're looking for.
SELECT
name,
id,
COUNT(id) as count
FROM
dataSet
WHERE
dataSet.name = 'input'
AND dataSet.id = 'input'
GROUP BY
name,
id
So I have a table like this:
Taxi Client Time
Tom A 1
Tom A 2
Tom B 3
Tom A 4
Tom A 5
Tom A 6
Tom B 7
Tom B 8
Bob A 1
Bob A 2
Bob A 3
and the expected result will be like this:
Tom 3
Bob 1
I have used the partition function to count the consecutive value but the result become this:
Tom A 2
Tom A 3
Tom B 2
Bob A 2
Please help, I am not good in English, thanks!
This is a variation of a gaps-and-islands problem. You can solve it using window functions:
select taxi, count(*)
from (select t.taxi, t.client, count(*) as num_times
from (select t.*,
row_number() over (partition by taxi order by time) as seqnum,
row_number() over (partition by taxi, client order by time) as seqnum_c
from t
) t
group by t.taxi, t.client, (seqnum - seqnum_c)
having count(*) >= 2
)
group by taxi;
use distinct count
select taxi ,count( distinct cient)
from table_name
group by taxi
It seems your expected output is wrong
I don't see where you get the number 3 from. If you're trying to do what your question says and group by client in consecutive order only and then get the number of different groups, I can help you out with the following query. Bob has 1 group and Tom has 4.
Partition by taxi, ORDER BY taxi, time and check if this client matches the previous client for this taxi. If yes, do not count this row. If no, count this row, this is a new group.
SELECT FEE.taxi,
SUM(FEE.clientNotSameAsPreviousInSequence)
FROM
(
SELECT taxi,
CASE
WHEN PreviousClient IS NULL THEN
1
WHEN PreviousClient <> client THEN
1
ELSE
0
END AS clientNotSameAsPreviousInSequence
FROM
(
SELECT *,
LAG(client) OVER (PARTITION BY taxi ORDER BY taxi, time) AS PreviousClient
FROM table
) taxisWithPreviousClient
) FEE
GROUP BY FEE.taxi;
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
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);
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