I hope you can help me with an SQL statment that I can't seem to figure out.
I have a table with a list of visits made, with the country visited and the dates such as;
United Kingdom, 1st Nov 2009
Germany, 8th June 2010
Frane, 10th September 2011
United Kingdom, 11th october 2011
etc.
I want to extract the data so that I get a table list such as follows
Times Visited - Country list
23 - United Kingdom
10 - France, Germany
4 - Czech Republic, USA, Canada
1 - Poland, Serbia, Argentina, New Zealand
So that the data shows that I have made 4 visist to the Czech Republic, USA & Canda
And the query
select * from
(
select count(*) as "trips", country from trip_view
group by country
order by 1 desc
) as a
returns the data
23 United Kingdom
10 France
10 Germany
4 Czech Republic
4 USA
4 Canada
etc.
So I need a kind of group on the outer SQL, but if I do the following
select * from
(
select count(*) as "trips", country from trip_view
group by country
order by 1 desc
) as a
group by trips
Then I only get 1 entry for each country. i.e.
23 United Kingdom
10 France
4 Czech Republic
etc.
So for each row I only have 1 country listed and not all of them. i.e. Row 2 should show France & Germany and row 3 should show Czech, USA & Canada
Any ideas how to do this in mysql?
Thanks
Using GROUP_CONCAT on the country column it should yield the result you want:
SELECT trips,
GROUP_CONCAT(country) AS country
FROM (
SELECT COUNT(*) AS "trips",
country
FROM trip_view
GROUP BY country
) a
GROUP BY trips
ORDER BY 1 DESC
Output:
TRIPS COUNTRY
23 United Kingdom
10 France,Germany
4 Czech Republic,Canada,USA
Live DEMO.
Related
How would I go about calculating the average and total Bonus from each City?
Example Input
Toronto 5
Toronto 8
New York 7
New York 3
London 10
Desired Output:
City Avg Total
Toronto 6.5 13
New York 5 10
London 10 10
This is a very simple group by statement:
SELECT City, AVG(Bonus) as `Avg`, SUM(Bonus) AS Total
FROM Cities
GROUP BY City
See example sqlfiddle here with your expected results.
You can do something like this:
SELECT city, AVG(bonus), SUM(bonus) FROM city_bonus GROUP BY city
Working fiddle
The table name is c_list.
Country City Rating Date
------------------------------------
France Brest 95 24092016
France Brest 98 27092016
France Brest 95 03102016
France Lille 100 26092016
France Lille 92 28092016
Japan Tokyo 98 02102016
There are more than 50 different countries and each country have few cities. And each city may have more than one row of record or more. I want to select one City with highest average Rating (Compare to Cities in it's own country) and then compare with all other cities in different countries. So, the final query should display all Country and their ONE City with max(avg(Rating)) and in desc order. The sample output:
Country City max(avg(rating))
-------------------------------------
USA New York 97.25
UK Cardiff 96.70
Germany Greven 96.50
Turkey Afyon 94.88
France Guipavas 94.10
Canada Cartwright 91.35
I can only get the max(avg(rating)) for one country. Need help.
SELECT top 1 country, city, Avg(rating) AS Ratings
FROM c_list
where country = 'France'
GROUP BY city, country
order by Ratings desc
(Edited) The result that I want is similar like Miss world contest. Compete and win against local contestant in your country first. Next (my final result set) is to compete against the winners from other countries and rank them first to last using their avg(rating) that they got eatlier in their country.
If am not wrong you are looking for this
SELECT country,
city,
Avg(rating) AS Ratings
FROM c_list A
GROUP BY city,
country
HAVING Avg(rating) = (SELECT TOP 1 Avg(rating) AS Ratings
FROM c_list B
WHERE a.country = b.country
GROUP BY city
ORDER BY ratings DESC)
ORDER BY ratings DESC
Note : If you are using Mysql the replace TOP keyword with LIMIT
Base table:
select t.* from temp_lax t
COUNTRY CITY RATING
1 France Brest 95
2 France Brest 98
3 France Brest 95
4 France Lille 100
5 France Lille 92
6 Japan Tokyo 98
Query:
select t1.country, t1.city, Avg(t1.Rating) rating
from temp_lax t1
group by t1.country, t1.city
having avg(t1.rating) = (select max(avg(rating))
from temp_lax t2
WHERE t1.country = t2.country
GROUP BY t2.city)
order by rating desc
OUTPUT:
COUNTRY CITY RATING
1 Japan Tokyo 98
2 France Lille 96
3 France Brest 96
Please let me know if you are looking for different result set.
So basically there are two tables.
First is location table -
id location_id city_name country_name
1 400212 Paris Canada
2 400122 Paris France
2 400122 Dubai United Arab Emirates
Second is general_details table -
id hotel_id city_name country_name
1 2323 Dubai United Arab Emirates
2 1231 Dubai United Arab Emirates
3 2344 Paris Canada
4 1218 Paris France
So lets suppose I have two country_name and city_name pairs from table locations. I want to select number of rows for them each from table general details. Like I have Paris => Canada and Paris => France, so my result set should have 1, 1 i.e. their respective total records from table general_details. I am not sure how to do this. I can have multiple pairs from table 1 and I want all the counts from table 2.
Like I made an array of city and country -
array[0][city] = Paris, array[0][country] = France
array[1][city] = Paris, array[1][country] = Canada
Now I want output resultset after 2nd query to have count(*) as 1,1.
Kindly guide me here.
Try This:
SELECT city_name,country_name, COUNT(*)
FROM
general_details
WHERE
city_name IN('Paris') AND country_name IN ('Canada' , 'France')
GROUP BY city_name,country_name
I'm new to SQL and my current sql query is:-
SELECT
CONCAT(LastName,', ',FirstName)AS 'Name',</li>
City,
Country,
ShipCity AS 'Shipped City'
From
itstudies.employees
INNER JOIN
Orders
ON Employees.EmployeeID = Orders.EmployeeID
WHERE City = ShipCity;
This query shows the output as:-
(Only part of the output is shown.)
Name City Country Shipped City
Smith, Jo York UK York
Avery, Paul Dallas USA Dallas
Avery, Paul Dallas USA Dallas
Kris, Jan York UK York
Kris, Jan York UK York
Hill, Ros Boston USA Boston
I need to take out the duplicates and change the query to show:-
Name City Country Shipped City
Smith, Jo York UK York
Avery, Paul Dallas USA Dallas
Kris, Jan York UK York
Hill, Ros Boston USA Boston
Thanks in advance.
You could just use
SELECT DISTINCT ...
but usually it is better to have a look at where the duplicates come from and work with a
GROUP BY
In this case you join with ORDERS and thus get duplicates for all orders of an employee. Is this join necessary? You could use
WHERE EXISTS (SELECT 1 FROM Orders WHERE Orders.EmployeeId=Employees.EmployeeId)
to get only employees who ordered.
Simple: Use the DISTINCT keyword.
You can refer to http://www.w3schools.com/sql/sql_distinct.asp for more details ;)
I have two tables on MySQL (using phpMyAdmin), looking like the following:
Table 1:
Country Total Minutes
USA 100
USA 90
Canada 60
Mexico 80
UK 90
France 70
France 10
Germany 10
In Table 2, what I need to do is the following:
Region Total Minutes
North America USA+USA+Canada+Mexico Mins
Europe UK+France+France+Germany Mins
Is there a way to have a row be the result of a query?
You either need a region column in table 1:
SELECT region, SUM(`Total Minutes`)
FROM timespent
GROUP BY region;
Or a separate region <-> country table:
SELECT region, SUM(`Total Minutes`)
FROM myregions r
INNER JOIN timespent t USING (country)
GROUP BY r.region;
The regions table would look like this:
region | country
--------------+--------
North America | USA
North America | Mexico
If you can't change anything in your database, look at Andomar's solution :)
You could translate the countries to regions in a subquery. The outer query can then group by on region:
select Region
, sum(TotalMinutes) as TotalMinutes
from (
select case country
when 'USA' then 'North America'
when 'France' then 'Europe'
end as Region
, TotalMinutes
from YourTable
) as SubQueryAlias
group by
Region