SQL Exercise: Sort the employees in ascending order on the salary
SQL employee Database: Exercise-74 with Solution
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74. From the following table, write a SQL query to list the employees in ascending order based on salary. Return complete information about the employees.
Sample table: employees
Sample Solution:
SELECT *
FROM employees
ORDER BY salary ASC;
Sample Output:
emp_id | emp_name | job_name | manager_id | hire_date | salary | commission | dep_id --------+----------+-----------+------------+------------+---------+------------+-------- 63679 | SANDRINE | CLERK | 69062 | 1990-12-18 | 900.00 | | 2001 69000 | JULIUS | CLERK | 66928 | 1991-12-03 | 1050.00 | | 3001 68736 | ADNRES | CLERK | 67858 | 1997-05-23 | 1200.00 | | 2001 65271 | WADE | SALESMAN | 66928 | 1991-02-22 | 1350.00 | 600.00 | 3001 66564 | MADDEN | SALESMAN | 66928 | 1991-09-28 | 1350.00 | 1500.00 | 3001 69324 | MARKER | CLERK | 67832 | 1992-01-23 | 1400.00 | | 1001 68454 | TUCKER | SALESMAN | 66928 | 1991-09-08 | 1600.00 | 0.00 | 3001 64989 | ADELYN | SALESMAN | 66928 | 1991-02-20 | 1700.00 | 400.00 | 3001 67832 | CLARE | MANAGER | 68319 | 1991-06-09 | 2550.00 | | 1001 66928 | BLAZE | MANAGER | 68319 | 1991-05-01 | 2750.00 | | 3001 65646 | JONAS | MANAGER | 68319 | 1991-04-02 | 2957.00 | | 2001 69062 | FRANK | ANALYST | 65646 | 1991-12-03 | 3100.00 | | 2001 67858 | SCARLET | ANALYST | 65646 | 1997-04-19 | 3100.00 | | 2001 68319 | KAYLING | PRESIDENT | | 1991-11-18 | 6000.00 | | 1001 (14 rows)
Explanation:
The said query in SQL that selects all columns from the 'employees' table and order the results by the "salary" column in ascending order.
This means that the query will return all employee records, but the records with the lowest salaries will appear first.
Relational Algebra Expression:

Relational Algebra Tree:

Practice Online
Sample Database: employee

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Previous SQL Exercise: Sort the employees by salary less than 1000.
Next SQL Exercise: Ascending by job name and descending by employee ID.
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SQL: Tips of the Day
Concatenate strings of a string field in a PostgreSQL 'group by' query:
Input:
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;
Output:
COMPANY_ID EMPLOYEE 1 Anna, Bill 2 Carol, Dave
Database: PostgreSQL
Ref: https://bit.ly/2XTiRjq
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