I am new to SQL (first time) and I try to get my result in 1 record:
I have a table with 3 entries: city_name, company_name, and employee number
Basically, in my table I have a company which repeats 2 times: city_name: Rome, company_name: ABC, employee number: 100 --> this entry is 2 times in the table
How can I make this 2 entries to be shown in the table as a single record and to sum my employee numbers. I need that 2x100 empolyee to be in a single shown as 200? So I will have something like this:
city_name company_name employee
Rome ABC 200
You are describing group by:
select city_name, company_name, sum(empoyee_number)
from t
group by city_name, company_name
Related
I am trying to design a database of customer details. Where customers can have up to two different phone numbers.
When I run the Select * command to bring out the customers that match criteria, I get this:
Name | Number
James | 12344532
James | 23232422
I would like it to display all customers with two numbers this way:
Name | Number | Number
James 12344532 23232422
John 32443322
Jude 12121212 23232422
I am using Postgresql server on Azure Data studio.
Please assist.
I tried using this command:
Select * FROM name.name,
min(details.number) AS number1,
max(details.number) AS number2
FROM name
JOIN details
ON name.id=details.id
GROUP BY name.name
I got this:
Name | Number | Number
James 12344532 23232422
John 32443322 32443322
Jude 12121212 23232422
Customers with just 1 phone number gets duplicated in the table. How do I go about this?
I would aggregate the numbers into an array, then extract the array elements:
select n.name,
d.numbers[1] as number_1,
d.numbers[2] as number_2
from name n
join (
select id, array_agg(number) as numbers
from details
group by id
) d on d.id = n.id
order by name;
This is also easy to extend if you have more than two numbers.
Try using the following query:
SELECT
Name,
MIN(CASE WHEN rn = 1 THEN Number END) AS Number1,
MIN(CASE WHEN rn = 2 THEN Number END) AS Number2
FROM
(SELECT
Name, Number,
ROW_NUMBER() OVER (PARTITION BY Name ORDER BY Number) AS rn
FROM name) t
GROUP BY Name
This query will use the ROW_NUMBER() function to assign a unique row number to each phone number for each customer. The ROW_NUMBER() function is ordered by the Number column, so the lowest number will have a row number of 1, and the second lowest number will have a row number of 2, etc.
Then we use the outer query to group customer by name and use MIN() function to get the first and second number based on the row number.
This query will return the desired output, with two columns, one showing the customer's first phone number and the other showing their second phone number.
Note: The query above assumes that the phone number is unique for each customer. If a customer has duplicate phone numbers, the query will return the first one it encounters.
This question already has answers here:
What's the SQL query to list all rows that have 2 column sub-rows as duplicates?
(10 answers)
Closed last year.
Suppose I have the following schema:
student(name, siblings)
The related table has names and siblings. Note the number of rows of the same name will appear the same number of times as the number of siblings an individual has. For instance, a table could be as follows:
Jack, Lucy
Jack, Tim
Meaning that Jack has Lucy and Tim as his siblings.
I want to identify an SQL query that reports the names of all students who have 2 or more siblings. My attempt is the following:
select name
from student
where count(name) >= 1;
I'm not sure I'm using count correctly in this SQL query. Can someone please help with identifying the correct SQL query for this?
You're almost there:
select name
from student
group by name
having count(*) > 1;
HAVING is a where clause that runs after grouping is done. In it you can use things that a grouping would make available (like counts and aggregations). By grouping on the name and counting (filtering for >1, if you want two or more, not >=1 because that would include 1) you get the names you want..
This will just deliver "Jack" as a single result (in the example data from the question). If you then want all the detail, like who Jack's siblings are, you can join your grouped, filtered list of names back to the table:
select *
from
student
INNER JOIN
(
select name
from student
group by name
having count(*) > 1
) morethanone ON morethanone.name = student.name
You can't avoid doing this "joining back" because the grouping has thrown the detail away in order to create the group. The only way to get the detail back is to take the name list the group gave you and use it to filter the original detail data again
Full disclosure; it's a bit of a lie to say "can't avoid doing this": SQL Server supports something called a window function, which will effectively perform a grouping in the background and join it back to the detail. Such a query would look like:
select student.*, count(*) over(partition by name) n
from student
And for a table like this:
jack, lucy
jack, tim
jane, bill
jane, fred
jane, tom
john, dave
It would produce:
jack, lucy, 2
jack, tim, 2
jane, bill, 3
jane, fred, 3
jane, tom, 3
john, dave, 1
The rows with jack would have 2 on because there are two jack rows. There are 3 janes, there is 1 john. You could then wrap all that in a subquery and filter for n > 1 which would remove john
select *
from
(
select student.*, count(*) over(partition by name) n
from student
) x
where x.n > 1
If SQL Server didn't have window functions, it would look more like:
select *
from
student
INNER JOIN
(
select name, count(*) as n
from student
group by name
) x ON x.name = student.name
The COUNT(*) OVER(PARTITION BY name) is like a mini "group by name and return the count, then auto join back to the main detail using the name as key" i.e. a short form of the latter query
You can do:
select name
from student as s1
where exists (
select s2
from student as s2
where s1.name = s2.name and s1.siblings != s2.siblings
)
I think the best approach is what 'Caius Jard' mentioned. However, additional way if you want to get how many siblings each name has .
SELECT name, COUNT(*) AS Occurrences
FROM student
GROUP BY name
HAVING (COUNT(*) > 1)
I wanted to share another solution I came up with:
select s1.name
from student s1, student s2
where s1.name = s2.name and s1.sibling != s2.sibling;
This is simple table and I have issue with group count.
STUID DEPT
100 Math
100 History
200 Math
200 History
This is my query
select count (distinct(STUID)), DEPT
from table
group by DEPT
and outcome count 2 for Math, 2 for History.
I need count without duplicate for STUID so the outcome look like
STUID DEPT
1 Math
1 History
My goal is to count stuid ( no dup ) group by Dept
I need count without duplicate for STUID
You query does that already: it counts each stuid just once, even if it appears multiple times for a given department. For example, if you had the following dataset, you would still get the same result (that is, a count of two for each department:
STUID DEPT
100 Math
100 History
200 Math
200 History
100 Math
100 History
200 Math
200 History
If you just want the distinct dept values, use distinct - we could add literal 1 to each row, but it wouldn't be very helfpul:
select distinct dept from mytable
I am not sure what title has to be here to correctly reflect my question, I can only describe what I want.
There is a table with fields:
id, name, city
There are next rows:
1 John London
2 Mary Paris
3 John Paris
4 Samy London
I want to get a such result:
London Paris
Total 2 2
John 1 1
Mary 0 1
Samy 1 0
So, I need to take all unique values of name and find an appropriate quantity for unique values of another field (city)
Also I want to get a total quantity of each city
Simple way to do it is:
1)Get a list of unique names
SELECT DISTINCT name FROM table
2)Get a list of unique cities
SELECT DISTINCT city FROM table
3)Create a query for every name and city
SELECT COUNT(city) FROM table WHERE name = some_name AND city = some_city
4)Get total:
SELECT COUNT(city) FROM table WHERE name = some_name
(I did't test these queries, so maybe there are some errors here but it's only to show the idea)
As there are 3 names and 2 cities -> 3 * 2 = 6 queries to DB
But for a table with 100 cities and 100 names -> 100 * 100 = 10 000 queries to DB
and it may take a lot of time to do.
Also, names and cities may be changed, so, I can't create a query with predefined names or cities as every day it's new ones, so, instead of London and Paris it may be Moscow, Turin and Berlin. The same thing with names.
How to get such table with one-two queries to original table using sqlite?
(sqlite: I do it for android)
You can get the per-name results with conditional aggregation. As for the total, unfortunately SQLite does not support the with rollup clause, that would generate it automatically.
One workaround is union all and an additional column for ordering:
select name, london, paris
from (
select name, sum(city = 'London') london, sum(city = 'Paris') paris, 1 prio
from mytable
group by name
union all
select 'Total', sum(city = 'London'), sum(city = 'Paris'), 0
from mytable
) t
order by prio, name
Actually the subquery might not be necessary:
select name, sum(city = 'London') london, sum(city = 'Paris') paris, 1 prio
from mytable
group by name
union all
select 'Total', sum(city = 'London'), sum(city = 'Paris'), 0
from mytable
order by prio, name
#GMB gave me the idea of using group by, but as I do it for SQLite on Android, so, the answer looks like:
SELECT name,
COUNT(CASE WHEN city = :london THEN 1 END) as countLondon,
COUNT(CASE WHEN city = :paris THEN 1 END) as countParis
FROM table2 GROUP BY name
where :london and :paris are passed params, and countLondon and countParis are fields of the response class
I wrote a query which need filter out the employee data on behalf on their employee codes.
For instance, in my XYZ table i have 200 employees, i need to insert these 200 employee in ABC table, but before inserting, i need to check whether all 200 employees are existed in the system,I first filter out the employee and then insert into my ABC table.
suppose, 180 out of 200 employee matched, then i will insert 180 in the ABC table.
Now i want the count 200-180=20, so i need that difference count.
I wrote a query but it fetches only the matched record, not those employee count who filters out.
Select distinct SD.EMP_code
FROm SALARY_DETAIL_REPORT_012018 SD /*219 Employees*/
JOIN
(SELECT * FROM EMPLOYEE) tbl
ON tbl.EMP_CODE=to_char(SD.EMP_CODE)
WHERE SD.REFERENCE_ID like '1-%';
final output : 213 employees
I want 219-213=6
i want those 6 employees. I also tried INTERSECT but i got same result.
Select distinct to_char(SD.EMP_code)
FROm SALARY_DETAIL_REPORT_012018 SD
WHERE SD.REFERENCE_ID like '1-%'
INTERSECT
SELECT EMP_CODE FROm EMPLOYEE;
OUTPUT
213 Employees
Kindly help me to find out the count of filtered employees
You can use NOT EXISTS :
SELECT DISTINCT SD.EMP_code
FROM SALARY_DETAIL_REPORT_012018 sd
WHERE NOT EXISTS (SELECT 1 FROM EMPLOYEE e WHERE e.EMP_CODE = TO_CHAR(SD.EMP_CODE)) AND
SD.REFERENCE_ID LIKE '1-%';
use except opertaor
Select distinct to_char(SD.EMP_code)
FROM SALARY_DETAIL_REPORT_012018 SD
WHERE SD.REFERENCE_ID like '1-%'
except
SELECT EMP_CODE FROm EMPLOYEE;