I am trying to get correct result but I can not find right way to get.
I have this table:
I would like to get result like this one
This is my code:
SELECT employee_ID, first_name, manager_id ,
(SELECT first_name from employees where MANAGER_ID= employee_id)
from employees
I got only first 3 columns and fourth one is empty because its wrong select
Do self join to the same table
select emp1.employee_ID, emp1.first_name, emp1.manager_id,emp2.first_name as manager_name from employees emp1
left join employees emp2
on emp1.manager_id=emp2.employee_id;
Sorry my bad explanation, I have 5 columns, first one in not in the picture because its only ID of row, my first row is Employee id, second one is first_name, third one is manager id and fourth one should be first name of manager. Manager name is that when manager id is equal to employee id, so e.g. employee id = 206, first name is William, manager id is 205. and name should be Shelley because Shelley's employee ID is 205
Related
select count(*),manager_id
from departments
group by manager_id;
this is my idea of how to, but it dosent gives me the amount of employees for each manager
I suspect you are selecting from the wrong table, the logic seems correct but the fact that you are selecting from a table call departments is a little suspicious.
Do you have an employee tables? Does it contains a manager_id column? If so:
select count(*),manager_id
from employees
group by manager_id;
If employee tables has only department_id column then :
SELECT d.manager_id,count(*)
FROM employees e
INNER JOIN departments d
ON(e.department_id = d.id)
Using the sample table from the HR schema
select MANAGER_ID, count(*), count(distinct EMPLOYEE_ID)
from HR.EMPLOYEES
group by MANAGER_ID
order by 1 nulls first;
gives
MANAGER_ID COUNT(*) COUNT(DISTINCTEMPLOYEE_ID)
---------- ---------- --------------------------
1 1
100 14 14
101 5 5
102 1 1
Note the first row, with manager IS NULL - i.e. there is a one employee without a manager.
Not also that I use both count(*)and count(distinct EMPLOYEE_ID). This is not relevant for this table, where EMPLOYEE_ID is PK, but in general case the former returns the number of record the latter the number of employees (which can be lower).
I have faced a similar problem of yours.
Try adding inside the count the id that you are counting ie employeeid.
If it still doesnt work after that try adding the same id in the group by.
Can you please help me with a query that would display a table like this:
Dept_ID Dept_Name
10 Admin
10 Whalen
20 Sales
20 James
20 King
20 Smith
40 Marketing
40 Neena
and so on...The Schema is HR
Display the Department Id and the Department Name and then the subsequent employees last names working under that department
SELECT Dept_ID, Dept_Name
FROM Your_Table
Simple as I can make it. It's very difficult (near impossible) to tell exactly what the query should be without more detail in terms of your table structure and some sample data.
From your edit, you may need something more like this;
SELECT DT.Dept_ID, DT.Dept_Name, ET.Emp_Name
FROM Dept_Table AS DT INNER JOIN Emp_Table AS ET ON DT.Dept_ID = ET.Dept_ID
ORDER BY Dept_ID
This shows the employees in each department on the next column, you don't really want all that in the same column.
When you union two data sets, there is NO implicit ordering, you could get the results in any order.
The get a particular order you must use ORDER BY.
To use ORDER BY, then you must have fields to do that ordering by.
In your case, the pseudo code would be...
- ORDER BY [dept_id], [depts-then-employees], [dept_name]
The middle of those three is something that YOU are going to have to create.
One way of doing that is as follows.
note: Just because you have a field to order by, does not mean that you have to select it.
SELECT
dept_id,
dept_name
FROM
(
SELECT
d.dept_id,
d.dept_name,
0 AS entity_type_ordinal
FROM
department d
UNION ALL
SELECT
d.dept_id,
e.employee_name,
1 AS entity_type_ordinal
FROM
department d
INNER JOIN
employee e
ON e.dept_id = d.dept_id
)
dept_and_emp
ORDER BY
dept_id,
entity_type_ordinal,
dept_name
Assuming there's a table in your database called departments that holds this information, your code might look like this:
select
dept_id, dept_name
from
departments
If you want to display certain columns of the table like you have asked in the question above , you can use the following syntax :
select column_names from table_name
replace:
column_names with the column names you want to display separated by a coma
2.table_name with the name of the table whose columns you wish to display
for the above question , the following code will do:
select Dept_Id , Dept_Name from Department ;
The above code works if your table name is 'Department'
SELECT
COUNT(emp.empNo)
FROM
Employee emp
WHERE
NOT EXISTS (SELECT dept.empNo
FROM department dept
WHERE emp.empNo = dept.empNo);
What does the where condition(where emp.empNo = dept.empNo) signify in the above query? I get different results with and without the where condition. I'm new to Oracle. Can any one help me to understand?
your query displaying count of employees which are not present in emp table but present in dept table.
suppose we have two tables emp and dept :
emp dept
1 1
2 2
3 3
4 4
5 5
6
7
from the given table we have emp 1 to 5 in both of the tables but in dept table having 2 employees (6,7)which are not present in emp table and your query is displaying count for those emp i.e 2
The query means that you're looking for only those employees, for which it does not exist a department with the same empNo as the employee's empNo.
I guess this is a query to find those employees which are not managers of any department (if we assume that the department's empNo is the empNo of the department's manager).
Still, it would be better if you provide the schema of the employee and the department tables.
The query is basically looking for the number of employees who don't belong to a department.
NOT EXISTS means that the query enclosed returns no rows. So, for any employee where no matching rows are found in the Department table they are counted.
Same as saying
SELECT
COUNT(emp.empNo)
FROM
Employee emp
WHERE
emp.EmpNo NOT IN ( SELECT
empNo
FROM
department)
I need to select 2 column from table with matched data from another table or cell be null ,
table 1 named "emp" contain emp_name ,emp_id
table 2 named "salary" contain emp_sal, emp_id
I need to create select query
have all emp_name, emp_id and emp_sal (for only the employees will take sale) or be Null
thanks for help now ((((update ))))
first thanks for help
i used
SELECT emp.emp_id,emp.emp_name,salary.emp_sal FROM emp LEFT JOIN salary ON emp.emp_id = salary.emp_id;
it work but with a lot of duplication and i need to make this query with day ...
i create another table named "day" i need query appear day i entered in this table
this table have only one column and i record ((day user entered and saved in "day.user_day"))
i need to link this three tables
together
and lets make it easy we will change salary to attendance ...
i need to query all names and id in date and apear all employee what ever thy have time or not
like when i search only in day 4/8/2014
name id time
john 1 04/08/2014 06:00
man 2 null
scsv 3 04/08/2014 07:00
You want a LEFT JOIN:
SELECT * FROM emp LEFT JOIN salary ON emp.emp_id = salary.emp_id;
This will return all employees along with their emp_sal, which will be NULL if it's not in the salary table.
If what I read is right, you want to get the employee name and salary, returning all employees regardless if they have an entry in salary. If that is correct, this should work:
SELECT
e.emp_name,
e.emp_id,
s.emp_sal
FROM
emp AS e
LEFT JOIN salary AS s ON e.emp_id = s.emp_id
If, however, you only wanted the employees with an entry in the salary table, change LEFT JOIN to be INNER JOIN.
I have an 'employee' table with
create table employee
(
e_number int,
e_name varchar(20),
salary money,
hire_date date
)
Now I want to display only the name of the employees who have the same name but different salary.
I tried select e_name,count(*) from employee group by e_name having count(*)>1;
but cannot combine it with "the same salary" section. Any help?
This assumes that you want the names of both of the people listed:
SELECT e1.e_name
FROM employee e1, employee e2
WHERE e1.e_name = e2.e_name
AND e1.salary <> e2.salary;
If you only want each name listed once, you would use a SELECT DISTINCT instead of the SELECT.
If your goal is to express this in the having clause:
Select name
from employee
group by name
having
count(*) > 1
and min(salary) != max(salary)
order by name
SELECT employee1.e_name, employee1.Salary, Employee2.Salary
FROM Employee employee1
JOIN Employee employee2
on employee1.name = employee2.name
AND Employee1.Salary <> Employee2.Salary
AND Employee1.E_Number <> employee2.E_Number
Basically get every employee, join it to every other employee via name, where the employee number is different (so don't join to yourself) and salary is different.
You probably don't need to check that employee number is different because 1 employee can only have 1 salary in your table design
Use a join, but importantly use a greater-than comparison, rather than a not-equals, to avoid duplicates:
SELECT e1.e_name as name1, e2.e_name as name2
FROM employee e1
JOIN employee e2 ON e1.e_name = e2.e_name
AND e1.salary > e2.salary;
Note also that the ON condition contains the salary comparison. It is a common misconception that the join on condition may only contain key-related comparisons. Doing this can have significant performance benefits, especially when further joins are made, because the ON condition is executed as the rows are joined - which discards non-matches immediately, whereas WHERE conditions are executed as a filter on the entire result set of the joins.
You just want the count of distinct salaries, not the count of all records.
select e_name,count(distinct salary)
from employee
group by e_name
having count(distinct salary)>1
(Drop the count in the select if unneeded - included since it was in your example)
First filter salary not double (not in), then grouping by e_name having count > 1
SELECT A.e_name
FROM employee A
WHERE A.salary NOT IN (SELECT salary FROM employee WHERE id != A.id)
GROUP BY A.e_name
HAVING COUNT(A.e_name) > 1