How to extract the higher values in sql query - sql

I have a question about sql. If i have a table with some information like that:
City Code
New York | 10A
New York | 10A
New York | 20A
L.A | 20A
L.A | 20A
Washington | 30A
Washington | 30A
Washington | 20A
San Francisco | 20A
San Francisco | 20A
with the query: SELECT City, Code, COUNT(Code) FROM table GROUP BY City I can get the following result:
City Code Total
New York | 10A 2
New York | 20A 1
L.A | 20A 2
Washington | 30A 2
Washington | 20A 1
San Francisco | 20A 2
The question is: How can I extract only the lines with higher total in each city?

Assuming SQL Server or a DBMS that has ranking functions:
SELECT *
FROM (SELECT City, Code,
COUNT(Code) Count_,
RANK() OVER (PARTITION BY City ORDER BY COUNT(*) DESC) Rank_
FROM #test
GROUP BY City,Code
) sub
WHERE Rank_ = 1 ;
This will return all tying entries, for example in case of a 3-way tie for Denver, you'd get the following:
DEN 10A 1, DEN 20A 1, Den 30A 1
Tested in SQL-Fiddle. Works in Oracle, Postgres, SQL-Server.

this worked for me:
SELECT y.city, z.code, z.total
FROM (SELECT city, MAX(total) AS total
FROM (SELECT city, code, COUNT(code) AS total
FROM tableName
GROUP BY city, code) AS x
GROUP BY city) AS y INNER JOIN
(SELECT city, code, COUNT(code) AS total
FROM tableName
GROUP BY city, code) AS z
ON y.city = z.city AND y.total = z.total
if there is a tie it will return all the rows that tied.

Related

How to aggregate using distinct values across two columns?

I have the following data in an orders table:
revenue expenses location_1 location_2
3 6 London New York
6 11 Paris Toronto
1 8 Houston Sydney
1 4 Chicago Los Angeles
2 5 New York London
7 11 New York Boston
4 6 Toronto Paris
5 11 Toronto New York
1 2 Los Angeles London
0 0 Mexico City London
I would like to create a result set that has 3 columns:
a list of the 10 DISTINCT city names
the sum of revenue for each city
the sum of expenses for each city
The desired result is:
location revenue expenses
London 6 13
New York 17 33
Paris 10 17
Toronto 15 28
Houston 1 8
Sydney 1 8
Chicago 1 4
Los Angeles 2 6
Boston 7 11
Mexico City 0 0
Is it possible to aggregate on distinct values across two columns? If yes, how would I do it?
Here is a fiddle:
http://sqlfiddle.com/#!9/0b1105/1
Shorter (and often faster):
SELECT location, sum(revenue) AS rev, sum(expenses) AS exp
FROM (
SELECT location_1 AS location, revenue, expenses FROM orders
UNION ALL
SELECT location_2 , revenue, expenses FROM orders
) sub
GROUP BY 1;
May be faster:
WITH cte AS (
SELECT location_1, location_2, revenue AS rev, expenses AS exp
FROM orders
)
SELECT location, sum(rev) AS rev, sum(exp) AS exp
FROM (
SELECT location_1 AS location, rev, exp FROM cte
UNION ALL
SELECT location_2 , rev, exp FROM cte
) sub
GROUP BY 1;
The (materialized!) CTE adds overhead, which may outweigh the benefit. Depends on many factors like total table size, available indexes, possible bloat, available RAM, storage speed, Postgres version, ...
fiddle
You could UNION ALL two queries and then select from it...
select location, sum(rev) as rev, sum(exp) as exp
from (
select location_1 as location, sum(revenue) as rev, sum(expenses) as exp
from orders
group by location_1
union all
select location_2 as location, sum(revenue) as rev, sum(expenses) as exp
from orders
group by location_2
)z
group by location
order by 1

Select unique countries with more than one customer

I need to show the countries that have more than one individual.
Customers
customer_id first_name last_name age country
1 John Doe 31 USA
2 Robert Luna 22 USA
3 David Robinson 22 UK
4 John Reinhardt 25 UK
5 Betty Doe 28 UAE
So the query should return
customer_id first_name last_name age country
1 John Doe 31 USA
2 Robert Luna 22 USA
3 David Robinson 22 UK
4 John Reinhardt 25 UK
I tried tis query but it didn't work.
SELECT last_name, Country
FROM Customers
GROUP BY Country
HAVING COUNT(Customer_id) > 1;
The actual table can be found here
Try using the following query. Thanks
SELECT * FROM CUSTOMERS C
WHERE C.COUNTRY IN (SELECT COUNTRY FROM CUSTOMERS GROUP BY COUNTRY HAVING COUNT(*)>1)
You could use a windowed count as a filter:
with c as (
select *, Count(*) over(partition by country) cnt
from Customers
)
select *
from c
where cnt > 1;

Display Value From Another Row

I have to write a query to display the value for ID2 when ID=ID. Currently the table looks like the below.
ID
ID2
fname
lname
address1
address2
city
state
123
123
joe
smith
12 main st
los angeles
CA
122
122
james
jones
13 main st
new york
NY
123
3210
joe
smith
14 main st
los angeles
CA
124
124
mary
jones
15 main st
new york
NY
The desired output would look like this. Where I can do some sort of a self join to get the ID2 value.
ID
ID2
fname
lname
address1
address2
city
state
other ID
123
123
joe
smith
12 main st
los angeles
CA
3210
122
122
james
jones
13 main st
new york
NY
124
124
mary
jones
15 main st
new york
NY
Any ideas/suggestions greatly appreciated!
Update:
I was able to get the result with the additional columns using the below script. Thanks all for your assistance on this one!
with cte (id, id2, fname, lname, address1, address2, city, state)
as (select *
from (select *,row_number() over(partition by id order by id desc) as rn from your_table
) x
where rn = 2)
select x.id, x.id2, x.fname, x.lname, x.address1, x.address2, x.city, x.state, c.id2, c.address1
from (select *,row_number() over(partition by id order by id desc) as rn from your_table
) x
left join
cte c on x.id=c.id
where x.rn = 1
dbfiddle below
https://dbfiddle.uk/?rdbms=postgres_9.5&fiddle=4010ab08c5e32d9293d10e985adbfd7a
You could do something like the following, using a correlated subquery; using max ensures it only returns a single row should you have an Id with more than one different ID2:
select *,
(select max(id2) from t t2 where t2.id = t.id and t2.id2 != t.id) OtherId
from t
where id = id2
If you have two rows max with the same id, then you can try this :
SELECT (array_agg(t.*) FILTER (WHERE id = id2))[1].*
, (array_agg(t.id2) FILTER (WHERE id <> id2))[1] AS "other ID"
FROM your_table AS t
GROUP BY id
If you may have more than two rows with the same id then you can try this :
SELECT (array_agg(t.*) FILTER (WHERE id = id2))[1].*
, array_agg(t.id2) FILTER (WHERE id <> id2) AS "other IDs"
FROM your_table AS t
GROUP BY id
see the test result in dbfiddle

Count occurrences with exclude criteria

I have a Table
City ID
Austin 123
Austin 123
Austin 123
Austin 145
Austin 145
Chicago 12
Chicago 12
Houston 24
Houston 45
Houston 45
Now I want to count the occurrences of all Citis with different ids so since Chicago has only one id (=12) I am not interested in Chicago and it should not appear in the resultset that should looks like this:
city Id Occurrences
Austin 123 3
Austin 145 2
Houston 34 1
Houston 45 2
I am able to get myself an overview with
select city, Id from Table
group by city, Id
But I am not sure how to only select the once having different ids and to count them.
Could anyone help me out here?
You can use window functions and aggregation:
select city, id, occurences
from (
select city, id, count(*) occurences, count(*) over(partition by city) cnt_city
from mytable
group by city, id
) t
where cnt_city > 1

SELECT Top 1 ID, DISTINCT Field

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