SQL Exercise: List id, name, department id, location of all employees
SQL employee Database: Exercise-93 with Solution
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93. From the following table, write a SQL query to list the employee id, name, department id, location of all the employees.
Sample table: employees
Sample table: department
SELECT e.emp_id, e.emp_name, e.dep_id, d.dep_location FROM employees e, department d WHERE e.dep_id = d.dep_id ;
emp_id | emp_name | dep_id | dep_location --------+----------+--------+-------------- 68319 | KAYLING | 1001 | SYDNEY 66928 | BLAZE | 3001 | PERTH 67832 | CLARE | 1001 | SYDNEY 65646 | JONAS | 2001 | MELBOURNE 67858 | SCARLET | 2001 | MELBOURNE 69062 | FRANK | 2001 | MELBOURNE 63679 | SANDRINE | 2001 | MELBOURNE 64989 | ADELYN | 3001 | PERTH 65271 | WADE | 3001 | PERTH 66564 | MADDEN | 3001 | PERTH 68454 | TUCKER | 3001 | PERTH 68736 | ADNRES | 2001 | MELBOURNE 69000 | JULIUS | 3001 | PERTH 69324 | MARKER | 1001 | SYDNEY (14 rows)
The said query in SQL that selects the emp_id, emp_name, dep_id from the employees table and the dep_location from the department table. The query joins the two tables on the dep_id column where they have the same values.
Relational Algebra Expression:
Relational Algebra Tree:
Sample Database: employee
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Previous SQL Exercise: Compute the total salary for each job in 1991.
Next SQL Exercise: List employee id of all the departments 1001 and 2001.
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SQL: Tips of the Day
Concatenate strings of a string field in a PostgreSQL 'group by' query:
ID COMPANY_ID EMPLOYEE 1 1 Anna 2 1 Bill 3 2 Carol 4 2 Dave
SELECT company_id, string_agg(employee, ', ') FROM mytable GROUP BY company_id;
COMPANY_ID EMPLOYEE 1 Anna, Bill 2 Carol, Dave
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