SELECT MIN(salary), MAX(salary) FROM employees;
SELECT team, MIN(salary), MAX(salary) FROM employees GROUP BY team;
USE world;
SELECT MAX(Population) AS 'Maximum Value'
FROM City
WHERE CountryCode = 'THA';
//returns the maximum (highest) value
SELECT MAX(working_hours) AS Maximum_working_hours FROM employee;
Code Example |
---|
Sql :: mysql case |
Sql :: sql display max value |
Sql :: how to delete all duplicate items in mysql |
Sql :: sql remove duplicates |
Sql :: show table info mysql |
Sql :: how to upper case in sql |
Sql :: creating table in mysql |
Sql :: sql get character at index |
Sql :: mysql_num_fields in mysqli |
Sql :: rollback in sql |
Sql :: delete rows from table sql |
Sql :: sql server 2019 installation |
Sql :: sql run multiple updates in one query |
Sql :: sql string function update replace |
Sql :: mysql table schema |
Sql :: orderBy sqlalchemy |
Sql :: sql restore backup query |
Sql :: drop database using terminal postgres |
Sql :: update select mysql |
Sql :: oracle generate list of dates in between a date range |
Sql :: mysql disable triggers |
Sql :: temp tables in sql server |
Sql :: mysql dump for selected row |
Sql :: start and stop mysql |
Sql :: sum sqlserver |
Sql :: remove root password mysql |
Sql :: sql group by example |
Sql :: export mysql table to file |
Sql :: Oracle filter date column by year |
Sql :: sql and |