I am having a question in SQL Joins. I have table employee with employeeid as primary key and some other columns for employee. And there is another table called employeeaddress where there can be multiple employeeid is a foreign key. One employee can have many employeeaddresses just to explain one to many relationship.
If I want to write a query which will fetch the following columns
employee.employeeid, employee.empname,
employeeaddress.employeeaddressid, employeeaddress.addr1,
employeeaddress.addr2
So there can be an employee with no employeeaddress. But anyway I wanted to fetch all the employees who may have zero or multiple addresses.
Do I need to apply left join or left outer join? I want the following result for a table that has 2 employees John and Michael where John has two employeeaddresses with employeeaddressid 21 and 22 and Michael has no employeeaddress
1, John, 21, addr1 for John, addr2 for John
1, John, 22, another addr1 for John, another addr2 for John
2, Michael, NULL , NULL , NULL
The above result is arranged in the following fashion
employee.employeeid, employee.empname, employeeaddress.employeeaddressid, employeeaddress.addr1, employeeaddress.addr2
Please help.
Based on your description it sounds like you're looking for a query as follows. If you also wanted the address details, you'll just have to add a left join to the outer query.
Also, as comments have eluded to, LEFT JOIN is shorthand for LEFT OUTER JOIN, they will produce the same results.
SELECT *
FROM employee
inner join
(
SELECT
employeeid,
count(*) as addresscount
FROM employee
left join employeeaddress ON employeeaddress.employeeaddressid = employee.employeeaddressid
group by employeeid
) counts on counts.employeeid = employee.employeeid
WHERE counts.addresscount = 0 -- Or 1, or 5 or > 1, etc.
LEFT JOIN should be all you need.
SQL Fiddle Example
SELECT e.employeeID ,
e.empName ,
ea.employeeAddressID ,
ea.addr1 ,
ea.addr2
FROM Employee e
LEFT JOIN EmployeeAddress ea ON ea.employeeID = e.employeeID
Related
I am learning postgresql and Inner join I have following table.
Employee
Id Name DepartmentId
1 John S. 1
2 Smith P. 1
3 Anil K. 2
Department
Department
Id Name
1 HR
2 Admin
I want to query to return the Department Name and numbers of employee in each department.
SELECT Department.name , COUNT(Employee.id) FROM Department INNER JOIN Employee ON Department.Id = Employee.DepartmentId Group BY Employee.department_id;
I dont know what I did wrong as I am new to database Query.
When involving all rows or major parts of the "many" table, it's typically faster to aggregate first and join later. Certainly the case here, since we are after counts for "each department", and there is no WHERE clause at all.
SELECT d.name, COALESCE(e.ct, 0) AS nr_employees
FROM department d
LEFT JOIN (
SELECT department_id AS id, count(*) AS ct
FROM employee
GROUP BY department_id
) e USING (id);
Also made it a LEFT [OUTER] JOIN, to keep departments without any employees in the result. And COALESCE to report 0 employees instead of NULL in that case.
Related, with more explanation:
Query with LEFT JOIN not returning rows for count of 0
Your original query would work too, after fixing the GROUP BY clause:
SELECT department.name, COUNT(employee.id)
FROM department
INNER JOIN employee ON department.id = employee.department_id
Group BY department.id; --!
That's assuming department.id is the PRIMARY KEY of the table, in which case it covers all columns of that table, including department.name. And you may want LEFT JOIN like above.
Aside: Consider legal, lower-case names exclusively in Postgres. See:
Are PostgreSQL column names case-sensitive?
I have three tables (example) STAFF, STU, EMP.
I want to combine the column EMPID in table STAFF and table EMP into 1 column?
My previous query is like this,
SELECT *
FROM STU s
FULL OUTER JOIN STAFF st ON st.STAFFID = STUID
FULL OUTER JOIN EMP e ON s.STUID = st.EMPID
The result is like this
The expected result is just like the above screenshot, but I want to join EMPID into one column only.
UPDATE:
I tried using this query:
SELECT
stu.stuid, stu.stuname, stu.stucode,
s.staffid, s.staffname, s.staffcode,
emp.empname, emp.empcode,
COALESCE (emp.empid, staff.staffid) AS col
FROM
STU, Staff, EMP
FULL OUTER JOIN
STAFF s ON s.STAFFID = stu.STUID
FULL OUTER JOIN
EMP e ON stu.STUID = s.EMPID
but it displays an error like this
Use below query to get the desired result.
SELECT s.StuID, s.StuName, s.Stucode, st.StaffId, st.StaffName, st.Staffcode, isnull(st.EmpId, e.EmpId) EmpId, e.EmpCode, e.EmpName
FROM STU s FULL outer JOIN
STAFF st
ON st.STAFFID = STUID FULL OUTER JOIN
EMP e
ON s.STUID = st.EMPID
Note: You will get the one emp Id column as needed. If Staff emp id is not null then staff emp id will be displayed else employee emp id will be displayed
I'm trying to accomplish something with my query but it's not really working. My application used to have a mongo db so the application is used to get arrays in a field, now we had to change to Postgres and I don't want to change my applications code to keep v1 working.
In order to get arrays in 1 field within Postgres I used array_agg() function. And this worked fine so far. However, I'm at a point where I need another array in a field from another different table.
For example:
I have my employees. employees have multiple address and have multiple workdays.
SELECT name, age, array_agg(ad.street) FROM employees e
JOIN address ad ON e.id = ad.employeeid
GROUP BY name, age
Now this worked fine for me, this would result in for example:
| name | age| array_agg(ad.street)
| peter | 25 | {1st street, 2nd street}|
Now I want to join another table for working days so I do:
SELECT name, age, array_agg(ad.street), arrag_agg(wd.day) FROM employees e
JOIN address ad ON e.id = ad.employeeid
JOIN workingdays wd ON e.id = wd.employeeid
GROUP BY name, age
This results in:
| peter | 25 | {1st street, 1st street, 1st street, 1st street, 1st street, 2nd street, 2nd street, 2nd street, 2nd street, 2nd street}| "{Monday,Tuesday,Wednesday,Thursday,Friday,Monday,Tuesday,Wednesday,Thursday,Friday}
But I need it to result:
| peter | 25 | {1st street, 2nd street}| {Monday,Tuesday,Wednesday,Thursday,Friday}
I understand it has to do with my joins, because of the multiple joins the rows multiple but I don't know how to accomplish this, can anyone give me the correct tip?
DISTINCT is often applied to repair queries that are rotten from the inside, and that's often expensive and / or incorrect. Don't multiply rows to begin with, then you don't have to fold unwanted duplicates at the end.
Joining to multiple n-tables ("has many") multiplies rows in the result set. That's efectively a CROSS JOIN or Cartesian product by proxy. See:
Two SQL LEFT JOINS produce incorrect result
There are various ways to avoid this mistake.
Aggregate first, join later
Technically, the query works as long as you join to one table with multiple rows at a time before you aggregate:
SELECT e.id, e.name, e.age, e.streets, array_agg(wd.day) AS days
FROM (
SELECT e.id, e.name, e.age, array_agg(ad.street) AS streets
FROM employees e
JOIN address ad ON ad.employeeid = e.id
GROUP BY e.id -- PK covers whole row
) e
JOIN workingdays wd ON wd.employeeid = e.id
GROUP BY e.id, e.name, e.age;
It's best to include the primary key id and GROUP BY it, because name and age are not necessarily unique. Else you might merge employees by mistake.
But better aggregate in a subquery before the join, that's superior without selective WHERE conditions on employees:
SELECT e.id, e.name, e.age, ad.streets, array_agg(wd.day) AS days
FROM employees e
JOIN (
SELECT employeeid, array_agg(ad.street) AS streets
FROM address
GROUP BY 1
) ad ON ad.employeeid = e.id
JOIN workingdays wd ON e.id = wd.employeeid
GROUP BY e.id, ad.streets;
Or aggregate both:
SELECT name, age, ad.streets, wd.days
FROM employees e
JOIN (
SELECT employeeid, array_agg(ad.street) AS streets
FROM address
GROUP BY 1
) ad ON ad.employeeid = e.id
JOIN (
SELECT employeeid, array_agg(wd.day) AS days
FROM workingdays
GROUP BY 1
) wd ON wd.employeeid = e.id;
The last one is typically faster if you retrieve all or most of the rows in the base tables.
Note that using JOIN and not LEFT JOIN removes employees from the result that have no row in address or none in workingdays. That may or may not be intended. Switch to LEFT JOIN to retain all employees in the result.
Correlated subqueries / JOIN LATERAL
For selective filters on employees, consider correlated subqueries instead:
SELECT name, age
, (SELECT array_agg(street) FROM address WHERE employeeid = e.id) AS streets
, (SELECT array_agg(day) FROM workingdays WHERE employeeid = e.id) AS days
FROM employees e
WHERE e.namer = 'peter'; -- very selective
Or LATERAL joins in Postgres 9.3 or later:
SELECT e.name, e.age, a.streets, w.days
FROM employees e
LEFT JOIN LATERAL (
SELECT array_agg(street) AS streets
FROM address
WHERE employeeid = e.id
GROUP BY 1
) a ON true
LEFT JOIN LATERAL (
SELECT array_agg(day) AS days
FROM workingdays
WHERE employeeid = e.id
GROUP BY 1
) w ON true
WHERE e.name = 'peter'; -- very selective
What is the difference between LATERAL JOIN and a subquery in PostgreSQL?
The last two queries retain all qualifying employees in the result.
Whenever you need values that aren't repeated, use DISTINCT, like so:
SELECT name, age, array_agg(DISTINCT ad.street), array_agg(DISTINCT wd.day) FROM employees e
JOIN address ad ON e.id = ad.employeeid
JOIN workingdays wd ON e.id = wd.employeeid
GROUP BY name, age
I am trying to convert the following query:
select *
from employees
where emp_id not in (select distinct emp_id from managers);
into a form where I represent the subquery as a join. I tried doing:
select *
from employees a, (select distinct emp_id from managers) b
where a.emp_id!=b.emp_id;
I also tried:
select *
from employees a, (select distinct emp_id from managers) b
where a.emp_id not in b.emp_id;
But it does not give the same result. I have tried the 'INNER JOIN' syntax as well, but to no avail. I have become frustrated with this seemingly simple problem. Any help would be appreciated.
Assume employee Data set of
Emp_ID
1
2
3
4
5
6
7
Assume Manger data set of
Emp_ID
1
2
3
4
5
8
9
select *
from employees
where emp_id not in (select distinct emp_id from managers);
The above isn't joining tables so no Cartesian product is generated... you just have 7 records you're looking at...
The above would result in 6 and 7 Why? only 6 and 7 from Employee Data isn't in the managers table. 8,9 in managers is ignored as you're only returning data from employee.
select *
from employees a, (select distinct emp_id from managers) b
where a.emp_id!=b.emp_id;
The above didnt' work because a Cartesian product is generated... All of Employee to all of Manager (assuming 7 records in each table 7*7=49)
so instead of just evaluating the employee data like you were in the first query. Now you also evaluate all managers to all employees
so Select * results in
1,1
1,2
1,3
1,4
1,5
1,8
1,9
2,1
2,2...
Less the where clause matches...
so 7*7-7 or 42. and while this may be the answer to the life universe and everything in it, it's not what you wanted.
I also tried:
select *
from employees a, (select distinct emp_id from managers) b
where a.emp_id not in b.emp_id;
Again a Cartesian... All of Employee to ALL OF Managers
So this is why a left join works
SELECT e.*
FROM employees e
LEFT OUTER JOIN managers m
on e.emp_id = m.emp_id
WHERE m.emp_id is null
This says join on ID first... so don't generate a Cartesian but actually join on a value to limit the results. but since it's a LEFT join return EVERYTHING from the LEFT table (employee) and only those that match from manager.
so in our example would be returned as e.emp_Di = m.Emp_ID
1,1
2,2
3,3
4,4
5,5
6,NULL
7,NULL
now the where clause so
6,Null
7,NULL are retained...
older ansii SQL standards for left joins would have been *= in the where clause...
select *
from employees a, managers b
where a.emp_id *= b.emp_id --I never remember if the * is the LEFT so it may be =*
and b.emp_ID is null;
But I find this notation harder to read as the join can get mixed in with the other limiting criteria...
Try this:
select e.*
from employees e
left join managers m on e.emp_id = m.emp_id
where m.emp_id is null
This will join the two tables. Then we discard all rows where we found a matching manager and are left with employees who aren't managers.
Your best bet would probably be a left join:
select
e.*
from employees e
left join managers m on e.emp_id = m.emp_id
where
m.emp_id is null;
The idea here is you're saying that you want to select everything from employees, including anything that matches in the manager table based on emp_id and then filtering out the rows that actually have something in the manager table.
Use Left Outer Join instead
select e.*
from employees e
left outer join managers m
on e.emp_id = m.emp_id
where m.emp_id is null
left outer join will preserve the rows from m table even if they do not have a match i e table based on the emp_id field. The we filter on where m.emp_id is null - give me all the rows from e where there's no matching record in m table.
A bit more on the subject can be found here:
Visual representation of joins
from employees a, (select distinct emp_id from managers) b implies cross join - all posible combinations between tables (and you needed left outer join instead)
The MINUS keyword should do the trick:
SELECT e.* FROM employees e
MINUS
Select m.* FROM managers m
Hope that helps...
select *
from employees
where Not (emp_id in (select distinct emp_id from managers));
I have two table one has employees goals and the other has list of employees. i have to match one to another. Seems easy to do. but in the employee table employees can be entered more than once with more than one way of spelling their names. How can I pick only one name for each ID, it really doesn't matter which one I pick.
this is the code i used:
select distinct (etar.EmplKey ), emp.EmplFullName
FROM EmployeeTarget etar
inner join DimEmployee emp on emp.emplkey = etar.emplkey
inner join dimbranch br on br.BranchId = etar.BranchId
where etar.BranchId = 8
this is the results i get:
EmplKey EmplFullName
100260 Ida Patton
101488 Don Sheppard
101488 Donald Sheppard
101489 Teresa Coverdale
103121 Harjinder Aujla
How can I have that Don Sheppard guy listed only once?
The easiest way is to do aggreagtion:
select etar.EmplKey, min(emp.EmplFullName)
FROM EmployeeTarget etar
inner join DimEmployee emp on emp.emplkey = etar.emplkey
inner join dimbranch br on br.BranchId = etar.BranchId
where etar.BranchId = 8
group by etar.EmplKey