Do I need to put all the column names in group by which I have select put in select?
for example in this simple query :
Select
CustomerID,
CompanyName,
ContactName,
ContactTitle,
City,
Country
From
Customers
Group By
Country,
CompanyName,
ContactName,
ContactTitle,
City,
Country,
CustomerID
I have to allways call same amount Group By what i used in Select?
If you're just selecting columns and you want the returned records to discard the exact duplicate rows? Then there are 2 methods.
1) group by
2) distinct
Your query doesn't use any of the aggregate functions like f.e. COUNT, MIN, MAX, SUM, ...
So your query could use DISTINCT instead of a GROUP BY.
select DISTINCT
CustomerID, CompanyName, ContactName, ContactTitle, City, Country
from Customers
But if CustomerID is a primary key, then CustomerID would already make the result unique.
So then this query doesn't need a GROUP BY or a DISTINCT to only get unique records.
select CustomerID, CompanyName, ContactName, ContactTitle, City, Country
from Customers
Note that one could have both DISTINCT and GROUP BY in the same query. But that's just pointless. A GROUP BY already enforces the uniqueness, so adding a DISTINCT to make them unique would just make the query slower for no reason.
As for the why all the columns in that select also have to be listed in the GROUP BY? Some databases, f.e. MySql can be more tolerant about not having to group on all columns. But it's a rule from one of the SQL Standards. So most databases enforce that. It's to avoid potential misleading results.
GROUP BY x, y means you want one result row per x and y. So if you have a table with bills, you could group by year and month for instance and thus get the number of bills (count(*)) and the total (sum(amount)) per month.
So the question is what rows do you want to see. A row per company (with the number of their customers) maybe? A row per city? The GROUP BY clause contains exactly those columns mentioned.
Your GROUP BY clause does exactly nothing, as select customers and you group by customer ID (which should be the customer table's primary key).
Related
I have a problem with the following task from the platform Codesignal:
After some investigation, you've created a database containing a foreignCompetitors table, which has the following structure:
competitor: the name of the competitor;
country: the country in which the competitor is operating.
In your report, you need to include the number of competitors per country and an additional row at the bottom that contains a summary: ("Total:", total_number_of_competitors)
Given the foreignCompetitors table, compose the resulting table with two columns: country and competitors. The first column should contain the country name, and the second column should contain the number of competitors in this country. The table should be sorted by the country names in ascending order. In addition, it should have an extra row at the bottom with the summary, as described above.
Example
For the following table foreignCompetitors
my solution:
CREATE PROCEDURE solution()
BEGIN
(SELECT country, COUNT(*) AS competitors
FROM foreignCompetitors
GROUP BY country
ORDER BY country)
UNION
SELECT 'Total:', COUNT(*) FROM foreignCompetitors;
END
But my output is:
The result of the countries is not sorted by their names.
I cannot understand why is that even though I try to sort them with ORDER BY.
You want a GROUP BY WITH ROLLUP here:
SELECT COALESCE(country, 'Total:') AS country, COUNT(*) AS competitors
FROM foreignCompetitors
GROUP BY country WITH ROLLUP
ORDER BY country;
If you want to stick with your union approach, then you need to introduce a computed column into the union query which places the total row at the bottom of the result set. Consider:
SELECT country, competitors
FROM
(
SELECT country, COUNT(*) AS competitors, 1 AS pos
FROM foreignCompetitors
GROUP BY country
UNION ALL
SELECT 'Total:', COUNT(*), 2
FROM foreignCompetitors
) t
ORDER BY pos, country;
So the goal is to get a list of customers that have on average ordered more than the total average of all customers.
Select customerNumber, customerName, orderNumber, SUM(quantityOrdered)as 'total_qty', ROUND(AVG(quantityOrdered),2) as 'avg'
From customers
join orders using(customerNumber)
join orderdetails using (orderNumber)
Group by customerNumber, OrderNumber
Having ROUND(AVG(quantityOrdered),2) > ROUND(AVG(quantityOrdered),2) IN
(SELECT ROUND(AVG(quantityOrdered),2) FROM orderdetails)
ORDER BY customerName;
My code runs but it doesn't filter the results on the avg quantity ordered column to only show results over the total average of 35.22.
Possibly, you mean:
select c.customernumber, c.customername,
sum(od.quantity_ordered) as sum_qty,
round(avg(od.quantity_ordered), 2) as avg_dty
from customers c
join orders o using(customerNumber)
join orderdetails od using (orderNumber)
group by customernumber, customername
having avg(od.quantity_ordered) > (select avg(quantity_ordered) from orderdetails)
Rationale:
you discuss computing the average ordered, but what your query does is compare the average order detail quantity per customer; this assumes that the latter is what you want
then: since you want an average per customer, so do not put the order number in the group by
no need for in or the-like in the having clause: just compare the customer's average against a scalar subquery that computes the overall
Notes:
don't use single quotes for identifiers (such as column aliases) - they are meant for literal strings
table aliases make the query easier to write and read; prefixing all columns with the alias of the table they belong to makes the query understandable
I have this table data:
I want to perform an sql query which will give me the total number of distinct loan applications per city.
So for example, I would expect this output
City Wexford
Loans 1
City Waterford 1
Loans 1
City Galway
Loans 3
Any idea what kind of query I need to perform to get the count of distinct loans for each city?
I would guess, probably a COUNT (Distinct ID) with GROUP BY. Something like this:
SELECT city, COUNT(DISTINCT LoanApplicationID) as Loans
FROM tableName
GROUP BY city
Here is another approach for this question. I am adding this because using DISTINCT may cause performance issues for another example, especially for large databases. Good to keep in mind.
select city,count(LoanApplicationID) as Loans
from (
select LoanApplicationID, city
from tablename
group by LoanApplicationID, city
) t
group by city
I have been at this for the past two hours and have tried many different ways in regards to subquery and joins. Here's the exact question "Get the name and city of customers who live in the city where the least number of products are made"
Here is a snapshot of the database tables
I know how to get the min
select min(quantity)
from products
but this returns just the min without the city attached to it so I can't search for the city in the customers table.
I have also tried group by and found it gave me 3 min's (one for each group of cities) which i believe may help me
select city,min(quantity)
from products
group by city
Putting everything together I got something that looks like
SELECT
c.name,c.city
FROM
customers c
INNER JOIN
(
SELECT
city,
MIN(quantity) AS min_quantity
FROM
products
GROUP BY
city
) AS SQ ON
SQ.city = c.city
But this returns multiple customers, which isn't correct. I assume by looking at the database the city when the lowest number of products seems to be Newark and there are no customers who reside in Newark so I assume again this query would result in 0 hits.Thank you for your time.
Example
Here is an example "Get the pids of products ordered through any agent who makes at least one order for a customer in Kyoto"
and the answer I provided is
select pid
from orders
inner join agents
on orders.aid = agents.aid
inner join customers
on customers.cid = orders.cid
where customers.city = 'Kyoto'
In Postgresql you have sophisticated tools, viz., windowing and CTEs.
WITH
find_least_sumq AS
(SELECT city, RANK() OVER ( PARTITION BY city ORDER BY SUM(quantity) ) AS r
FROM products)
SELECT name, city
FROM customers NATURAL JOIN find_least_sumq /* ON city */
WHERE r=1; /* rank 1 is smallest summed quantity including ties */
In Drew's answer, you are zeronig in on the cities where the smallest number of any particular item is made. I interpret the question as wanting the sum of items made in that city.
I guess it be something around this idea:
select customers.name, city.city, city.min
from customers
join (
select city, sum (quantity) as min
from products
group by city
--filter by the cities where the total_quantity = min_quantity
having sum (quantity) = (
--get the minimum quantity
select min(quantity) from products
)
) city on customers.city = city.city
This can be made so much simpler. Just sort the output by the field you want to get the minimum of.
SELECT city, quantity FROM customers ORDER BY quantity LIMIT 1;
I have just figured out my own answer. I guess taking a break and coming back to it was all I needed. For future readers this answer will use a subquery to help you get the min of a column and compare a different column (of that same row) to a different tables column.
This example is getting the city where the least number of products are made (quantity column) in the products table and comparing that city to the cities to the city column in the customers table, then printing the names and the city of those customers. (to help clarify, use the link in the original question to look at the structure of the database I am talking about) First step is to sum all the products to their respective cities, and then take the min of that, and then find the customers in that city.Here was my solution
with citySum as(
select city,sum(quantity) as sum
from products
group by city)
select name,city
from customers
where city
in
(select city
from citySum
where sum =(
select min(sum)
from citySum))
Here is another solution I have found today that works as well using only Sub queries
select c.name,c.city
from customers c
where c.city
in
(select city
from
(select p.city,sum(p.quantity) as lowestSum
from products p
group by p.city) summedCityQuantities
order by lowestsum asc
limit 1)
I have a table of users, and in this table I have a country field telling where these people are from (i.e. "Sweden", "Italy", ...). How can I do a SQL query to get something like:
Country Number
Sweden 10
Italy 50
... ...
Users select their countries from a list I give to them, but the list is really huge so it would be great to have a SQL query that can avoid using that list, that is look in the DB and give back only those countries which are in the database, because for example I have nobody from Barbados, even if I have that option in the country select field of the signup form :)
Thanks in advance!
If the name of the country is in the Users table, try something like this:
SELECT Country, COUNT (*) AS Number
FROM Users
GROUP BY Country
ORDER BY Country
If the name of the country is in the country table, then you will have to join
SELECT Contries.CountryName, Count (*) AS Number
FROM Users
INNER JOIN Countries
ON Users.CountryId = Countries.CountryId
GROUP BY Countries.CountryName
ORDER BY Countries.CountryName
This will give what you want. But you might want to cache the result of the query. With a lot of users it's quite a heavy query.
SELECT
country,
COUNT(*)
FROM
users
GROUP BY
country
Perhaps a better idea is (assuming you don't need the counts) to do it like this:
SELECT
DISTINCT country
FROM
users
Sounds like you want something like this...?
SELECT Country, COUNT(*) AS Number
FROM Users
GROUP BY Country
This is pretty straightforward:
SELECT
Country, COUNT(*) AS 'Number'
FROM
YourTable
GROUP BY
Country
ORDER BY
Country
You just group your data by country and count the entries for each country.
Or if you want them sorted by the number of visitors, use a different ORDER BY clause:
SELECT
Country, COUNT(*) AS 'Number'
FROM
YourTable
GROUP BY
Country
ORDER BY
COUNT(*) DESC
If you want the count per country:
select country, count(*) from users group by country;
If you just want the possible values:
select distinct country from users;
SELECT BillingCountry, COUNT(*)as Invoices
FROM Invoice
GROUP BY BillingCountry
ORDER BY Invoices DESC