SQL exercises on employee Database: Display all the information of the employees
SQL employee Database: Exercise-1 with Solution
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1. From the following table return complete information about the employees.
Sample table: employees
Pictorial Presentation:
Sample Solution:
SELECT * FROM employees;
Sample Output:
emp_id | emp_name | job_name | manager_id | hire_date | salary | commission | dep_id --------+----------+-----------+------------+------------+---------+------------+-------- 68319 | KAYLING | PRESIDENT | | 1991-11-18 | 6000.00 | | 1001 66928 | BLAZE | MANAGER | 68319 | 1991-05-01 | 2750.00 | | 3001 67832 | CLARE | MANAGER | 68319 | 1991-06-09 | 2550.00 | | 1001 65646 | JONAS | MANAGER | 68319 | 1991-04-02 | 2957.00 | | 2001 67858 | SCARLET | ANALYST | 65646 | 1997-04-19 | 3100.00 | | 2001 69062 | FRANK | ANALYST | 65646 | 1991-12-03 | 3100.00 | | 2001 63679 | SANDRINE | CLERK | 69062 | 1990-12-18 | 900.00 | | 2001 64989 | ADELYN | SALESMAN | 66928 | 1991-02-20 | 1700.00 | 400.00 | 3001 65271 | WADE | SALESMAN | 66928 | 1991-02-22 | 1350.00 | 600.00 | 3001 66564 | MADDEN | SALESMAN | 66928 | 1991-09-28 | 1350.00 | 1500.00 | 3001 68454 | TUCKER | SALESMAN | 66928 | 1991-09-08 | 1600.00 | 0.00 | 3001 68736 | ADNRES | CLERK | 67858 | 1997-05-23 | 1200.00 | | 2001 69000 | JULIUS | CLERK | 66928 | 1991-12-03 | 1050.00 | | 3001 69324 | MARKER | CLERK | 67832 | 1992-01-23 | 1400.00 | | 1001 (14 rows)
Relational Algebra Expression:
Relational Algebra Tree:
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Sample Database: employees
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Next: From the following table, write a SQL query to find the salaries of all employees. Return salary.
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SQL: Tips of the Day
SQL Server SELECT into existing table.
INSERT INTO dbo.TABLETWO SELECT col1, col2 FROM dbo.TABLEONE WHERE col3 LIKE @search_key
This assumes there's only two columns in dbo.TABLETWO - you need to specify the columns otherwise:
INSERT INTO dbo.TABLETWO (col1, col2) SELECT col1, col2 FROM dbo.TABLEONE WHERE col3 LIKE @search_key
Database: SQL Server
Ref: https://bit.ly/3y6tpA3
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