I have to make a query that has the Total number of customers by country and city
country and city are columns that are inside the customer table
On my own I have managed to get the total number of customers per city like this:
SELECT city, COUNT (*)
FROM employees
GROUP BY city
ORDER BY city
But how do I get it together with the country?
looking for information I think it should be something like this and ordered from largest to smallest
Country
City
TOTAL_CUSTOMERS
USA
Kirkland
3
USA
London
2
UK
Redmond
2
UK
Seattle
1
UK
Tacoma
1
What we have been told is to say Total number of customers by country and city.
You simply add country to the column list and group by list:
SELECT country,city, COUNT(*)
FROM employees
GROUP BY country,city
ORDER BY COUNT(*) DESC
Related
I have 3 different tables, country, city, and customer. Tables are shown below:
country table:
id country_name
1 UK
2 US
3 Brazil
:
n Canada
city table
id city_name postal_code country_id
1 London 30090 1
2 Dallas 20909 2
3 Rio 29090 3
4 Atlanta 30318 2
:
n Vancouver 32230 n
customer table
id customer_name city_id
1 John 1
2 Pete 3
3 Dave 2
4 May 2
5 Chuck 4
6 Sam 3
7 Henry 3
***country.id is references city.country_id, and city.id is references customer.city_id
I want to write a query that can extract the country name, city name and the count of the customer of the associate city. But with one condition, the query will return all cities with more customers than the average number of customers of all cities
It will look something like below, this is the correct output
UK London 2
Brazil Rio 3
but I kept getting this output, which isn't correct
UK London 2
US Dallas 2
US Atlanta 1
Brazil Rio 3
I fixed my SQL query but it doesn't give me the result that I want
SELECT country.country_name, city.city_name, COUNT(customer.city_id) from country
JOIN city on country.id = city.country_id
JOIN customer on city.id = customer.city_id
Group by city_name,country.country_name;
I am wondering how can I do this and fix my code?
add country.country_name in group by
SELECT country.country_name, city.city_name, COUNT(customer.city_id) from country
JOIN city on country.id = city.country_id
JOIN customer on city.id = customer.city_id
Group by city_name,country.country_name
You are missing country.country_name in the query it will give you error. As a general rule all columns on which aggregate function is not applied should be part of group by clause.
So either you write your query without country_name in the select or add country_name in the group by clause.
The question asked to take the city with the highest sum of goods that bought by customer for each country. Basically, there are cities that have the same number of goods, but we only keep the first one in alphabetical order. The result only contains country name, the city with highest number of goods and their goods in sum.
Table Schema:
Country table:
country_name
city_name
Goods table:
city_name
user_id
number_of_goods
My queries result:
France Paris 85
Germany Berlin 100
Germany Frankfurt 100
Germany Luxembourg 100
Netherlands Amsterdam 75
Spain Barcelona 93
The right result should be:
France Paris 85
Germany Berlin 100
Netherlands Amsterdam 75
Spain Barcelona 93
You can use row_number() :
select t.*
from (select t.*, row_number() over (partition by country order by city) as seq,
max(no_goods) over (partition by country) as max_good
from table t
) t
where seq = 1;
use aggregation functions min() for city and max() for no_of_goods.
select t1.country, t1.no_of_goods, min(t2.city) as city
from
(select country, max(no_of_goods) as no_of_goods from tableA
group by country) t1
left join tableA t2 on t2.no_of_goods = t1.no_of_goods and t1.country = t2.country
group by t1.country, t1.no_of_goods
see dbfiddle.
Basically, there are cities that have the same number of goods, but we only keep the first one in alphabetical order.
Based on your sample data, all cities in a country seem to have the same number_of_goods. If so, you can just use aggregation:
select c.country, min(c.city_name), max(number_of_goods)
from countries c join
goods g
on c.city_name = g.city_name
group by c.country;
So I have this table of subscribers of users and the country they are in.
UserID | Name | Country
-------+-------------------+------------
1 | Zaphod Beeblebrox | UK
2 | Arthur Dent | UK
3 | Gene Kelly | USA
4 | Nat King Cole | USA
I need to produce a list of all the users by percentage from each of the countries. I also need all the smaller member countries (under 1%) to be collapsed into an "OTHERS" category.
I can accomplish a simple "top x" of members trivially with a
SELECT COUNTRY, COUNT(*) AS POPULATION FROM SUBSCRIBERS GROUP BY COUNTRY ORDER BY POPULATION DESC LIMIT 10
and can generate the percentages by PHP server side code, but I don't quite know how to:
Do all of it in SQL including percentage calculations directly in the result
Club all under 1% members into a single OTHERS category.
So I need something like this:
Country | Population
--------+-----------
USA | 25.4%
Brazil | 12%
UK | 5%
OTHERS | 65%
Appreciate the help!
Here is query for this, I used a subquery to count the total number of rows and then used that to get the percentage value for each. The 'Others' category was generated in a separate query. Rows are sorted by descending population with the Others row last.
SELECT * FROM
(SELECT country , ROUND((100.0*COUNT(*)/count_all),1) ||'%' AS population
FROM (SELECT count(*) count_all FROM subscribers) AS sq,
subscribers s
WHERE (SELECT 100*count(*)/count_all
FROM subscribers s2
WHERE s2.country = s.country) > 1
GROUP BY country
ORDER BY population DESC)
UNION ALL
SELECT 'OTHERS', IFNULL(ROUND(100.0*COUNT(*)/count_all,1),0.0) ||'%' AS population
FROM (SELECT count(*) count_all FROM subscribers) AS sq,
subscribers s
WHERE (SELECT 100*count(*)/count_all
FROM subscribers s2
WHERE s2.country = s.country) <= 1
Ok I think I might have found a way to do this that's a hell of a lot quicker on execution speed:
SELECT territory,
Round(Sum(percentage), 3) AS Population
FROM (SELECT
Round((Count(*)*100.0)/(SELECT Count(*) FROM subscribers),3) AS Percentage,
CASE
WHEN ((Count(*)*100.0)/(SELECT Count(*) FROM subscribers)) > 2 THEN
country
ELSE 'Other'
END AS Territory
FROM subscribers
GROUP BY country
ORDER BY percentage DESC)
GROUP BY territory
ORDER BY population DESC;
Example in my employee table i having these data
//EMPLOYEE
E# NAME CITY
--------------------------
1 JOHN ENGLAND
2 SITI ENGLAND
3 JESS GERMANY
4 HOLY BOSTON
When i run this statement:
SELECT DISTINCT CITY, COUNT(E#) as "Total Employee" FROM EMPLOYEE Where CITY=;
what should i insert the value for
//CITY=?
in order to get the result like this
CITY TOTAL EMPLOYEE
----------------------------
ENGLAND 2
GERMANY 1
BOSTON 1
isn't impossible to use group by or having clause?
ANSWER FIXED ! THANKS
Well, you're not excluding any cities, so you don't need a WHERE clause at all! But you DO need a GROUP BY to do a count. You can just do:
SELECT CITY, COUNT(E#) as "Total Employee"
FROM EMPLOYEE
GROUP BY CITY
However if you only want to include those three cities (even if more are added to the underlying data), you can do:
SELECT CITY, COUNT(E#) as "Total Employee"
FROM EMPLOYEE
WHERE CITY IN ('ENGLAND', 'GERMANY', 'BOSTON')
GROUP BY CITY
I have a table with customer information like date of birth, address, contactinfo, etc.
I want to count the number of customers per city with a single query that outputs two values per record, cityname and amount of customers living there:
Alabama 285
Kentucky 167
New York 4
Rio de Janeiro 950
etc...
There are hundreds of cities in the table so I can't do a
SELECT count(CASE WHEN city = 'Alabama' THEN 1 END) AS Alabama
You can use GROUP BY clause to count the number of customers per city :
SELECT city
, COUNT(*)
FROM table
GROUP BY city