This simple query satisfy me
db2 => SELECT city,SUM(sales) as sum from offices group by city;
CITY SUM
---------------------------------------------------------------------------------------------------- ---------------------------------
Rome 14000,
Paris 19000,
But..how to add the $ or € symbol? To get an output like this?
CITY SUM
---------------------------------------------------------------------------------------------------- ---------------------------------
Rome 14000$
Paris 19000$
Normally, this kind of formatting is left to the client to display.
If you really need to have the DB do it, then VARCHAR_FORMAT is likely a better choice.
select city
, varchar_format(sum(sales),'$999G999G990D99' as sales
from offices group by city;
The G and D characters represent the grouping and decimal character for your locale
Solution found using the ||
select city,sum(sales) || '$' as sales from offices group by city;
Related
I am writing an SQL query to select each row except those which have mentioned city values.
But somehow it's not working and selects rows with these cities.
Select * from Emp
Where City not in ('Suart', 'Vapi');
Upcoming output
Name City
Kris Surat
Joy vapi
Riva Goa
Jeni Mumbai
Maya Sayan
Expected output
Name City
Riva Goa
Jeni Mumbai
Maya Sayan
Maybe your data must be case sensitive or spaces added.
Try this if it works
Select * from Emp
Where lower(ltrim(rtrim(City))) not in ('suart', 'vapi');
Or
Select * from Emp
Where lower(ltrim(rtrim(City))) not like '%suart%' OR
lower(ltrim(rtrim(City))) not like '%vapi%'
We are learning SQL at school, and my professor has this sql code in his documents.
SELECT wp.city, (COUNT(*) / locations.area) AS population_density
FROM world_poulation AS wp
INNER JOIN location
ON wp.city = locations.city
WHERE locations.state = “Hessen”
GROUP BY wp.city, locations.area
Everything is almost clear for me, just the aggregate function with /locations.area doesn't make any sense to me. Can anybody help?
Thank you in advance!
Look at what the query is grouped on, that tells you what each group consists of. In this case, each group is a city, and contains all the rows that have the same value for wp.city (and as the location table is joined on that value too, the locations.area is only included in the grouping so that it can be used in the result).
So each group has a number of rows, and the COUNT(*) aggregate will contain the number of rows for each group. The value of (COUNT(*) / locations.area) will be the number of rows in the group divided by the value of locations.area for that group.
If you would have data like this:
world_population
name city
--------- ---------
John London
Peter London
Sarah London
Malcolm London
Ian Cardiff
Johanna Stockholm
Sven Stockholm
Egil Stockholm
locations
city state area
----------- -------------- ---------
London Hessen 2
Cardiff Somehere else 14
Stockholm Hessen 1
Then you would get a result with two groups (as Cardiff is not in the state Hessen). One group has four people from London which has the area 2, so the population density would be 2. The other group has three people from Stockholm which has the area 1, so the population density would be 3.
Side note: There is a typo in the query, as it joins in the table location but refers to it as locations everywhere else.
Try writing it like:
SELECT wp.city,
locations.area,
COUNT(*) AS population,
(COUNT(*) / locations.area) AS population_density
FROM world_poulation AS wp
INNER JOIN location
ON wp.city = locations.city
WHERE locations.state = “Hessen”
GROUP BY wp.city, locations.area
The key is the GROUP BY statement. You are showing pairs of cities and areas. The COUNT(*) is the number of times a given pair shows up in the table you created by joining world population and location. The area is just a number, so you can divide the area by the COUNT.
I am looking at making a simple leader board for a time trial. A member may perform many time trials, but I only want for their fastest result to be displayed. My table columns are as follows:
Members { ID (PK), Forename, Surname }
TimeTrials { ID (PK), MemberID, Date, Time, Distance }
An example dataset would be:
Forename | Surname | Date | Time | Distance
Bill Smith 01-01-11 1.14 100
Dave Jones 04-09-11 2.33 100
Bill Smith 02-03-11 1.1 100
My resulting answer from the example above would be:
Forename | Surname | Date | Time | Distance
Bill Smith 02-03-11 1.1 100
Dave Jones 04-09-11 2.33 100
I have this so far, but access complains that I am not using Date as part of an aggregate function:
SELECT Members.Forename, Members.Surname, Min(TimeTrials.Time) AS MinOfTime, TimeTrials.Date
FROM Members
INNER JOIN TimeTrials ON Members.ID = TimeTrials.Member
GROUP BY Members.Forename, Members.Surname, TimeTrials.Distance
HAVING TimeTrials.Distance = 100
ORDER BY MIN(TimeTrials.Time);
IF I remove the Date from the SELECT the query works (without the date). I have tried using FIRST upon the TimeTrials.Date, but that will return the first date which is normally incorrect.
Obviously putting the Date as part of the GROUP BY would not return the result set that I am after.
Make this task easier on yourself by starting with a smaller piece of the problem. First get the minimum Time from TimeTrials for each combination of MemberID and Distance.
SELECT
tt.MemberID,
tt.Distance,
Min(tt.Time) AS MinOfTime
FROM TimeTrials AS tt
GROUP BY
tt.MemberID,
tt.Distance;
Assuming that SQL is correct, use it in a subquery which you join back to TimeTrials again.
SELECT tt2.*
FROM
TimeTrials AS tt2
INNER JOIN
(
SELECT
tt.MemberID,
tt.Distance,
Min(tt.Time) AS MinOfTime
FROM TimeTrials AS tt
GROUP BY
tt.MemberID,
tt.Distance
) AS sub
ON
tt2.MemberID = sub.MemberID
AND tt2.Distance = sub.Distance
AND tt2.Time = sub.MinOfTime
WHERE tt2.Distance = 100
ORDER BY tt2.Time;
Finally, you can join that query to Members to get Forename and Surname. Your question shows you already know how to do that, so I'll leave it for you. :-)
I have a table similar to the below:
id | name | direction |
--------------------------------------
1 Jhon Washington, DC
2 Diego Miami, Florida
3 Michael Orlando, Florida
4 Jenny Olympia, washington
5 Joe Austin, Texas
6 Barack Denver, Colorado
and I want to count how many people live in a specific state:
Washington 2
Florida 2
Texas 1
Colorado 1
How can I do this? (By the way this is just an question with an academic point of view )
Thanks in advance!
Postgres offers the function split_part(), which will break up a string by a delimiter. You want the second part (the part after the comma):
select split_part(direction, ', ', 2) as state, count(*)
from t
group by split_part(direction, ', ', 2);
Initially I would obtain the state from the direction field. Once you have that, it's quite simple:
SELECT state, count(*) as total FROM initial_table group by state.
To obtain the state, some functions depending on the dbms are useful. It depends on the language.
A possible pseudocode (given a function like substring_index of MySQL) for the query would be:
SELECT substring_index(direction,',',-1) as state, count(*) as total
FROM initial_table group by substring_index(direction,',',-1)
Edit: As it is suggested above, the query should return 1 for the Washington state.
My way do making such a queries is two-step - first, prepare fields you need, second, do you grouping or other calculation. That way you're following DRY principle and don't repeating yourself. I think CTE is the best tool for this:
with cte as (
-- we don't need other fields, only state
select
split_part(direction, ', ', 2) as state
from table1
)
select state, count(*)
from cte
group by state
sql fiddle demo
If you writing queries that way, it's easy to change grouping field in the future.
Hope that helps, and remember - readability counts! :)
I have problem with sql query.
For example, I have table like this:
ID Name1 Name2 Country
1 Greg Torr Poland
2 John Smith England
3 Tom Jerry USA
I want get all record, which have for example, "la" in Country. In this case:
PoLAnd
EngLAnd
How I can put this in Where clausule?
Greets
Use the LIKE keyword:
SELECT * FROM table
WHERE Country LIKE '%la%'
I think you can use the Like Clause here. The examples are available on
http://www.sql-tutorial.net/SQL-LIKE.asp
http://www.w3schools.com/sql/sql_like.asp
etc.
WHERE CHARINDEX('LA', Country) > 0
alternatively
WHERE Country LIKE '%la%'
If your RDBMS is case-sensitive, convert Country to upper case using the appropriate string function, and compare against the upper case LA with a LIKE:
SELECT *
FROM tbl
WHERE UPPER(Country) LIKE '%LA%'
Enclose your keyword before and after with Percent Symbol
SELECT *
FROM tableName
WHERE Country LIKE '%LA%'