/*mysql> select * from employee;
+----+-----------+----------+----------------------------+------+---------------+--------+-------+---------------------+
| id | firstname | lastname | title | age | yearofservice| salary | perks | email |
+----+-----------+----------+----------------------------+------+---------------+--------+-------+---------------------+
| 1 | John | Chen | Senior Programmer | 31 | 3| 120000 | 25000 | j@hotmail.com |
| 2 | Jan | Pillai | Senior Programmer | 32 | 4| 110000 | 20000 | g@yahoo.com |
| 3 | Ane | Pandit | Web Designer | 24 | 3| 90000 | 15000 | a@gmail.com |
| 4 | Mary | Anchor | Web Designer | 27 | 2| 85000 | 15000 | m@mail.com |
| 5 | Fred | King | Programmer | 32 | 3| 75000 | 15000 | f@net.com |
| 6 | John | Mac | Programmer | 32 | 4| 80000 | 16000 | j@hotmail.com |
| 7 | Arthur | Sam | Programmer | 28 | 2| 75000 | 14000 | e@yahoo.com |
| 8 | Alok | Nanda | Programmer | 32 | 3| 70000 | 10000 | a@yahoo.com |
| 9 | Susan | Ra | Multimedia Programmer | 32 | 4| 90000 | 15000 | h@gmail.com |
| 10 | Paul | Simon | Multimedia Programmer | 23 | 1| 85000 | 12000 | ps@gmail.com |
| 11 | Edward | Parhar | Multimedia Programmer | 30 | 2| 75000 | 15000 | a@hotmail.com |
| 12 | Kim | Hunter | Senior Web Designer | 32 | 4| 110000 | 20000 | kim@coolmail.com |
| 13 | Roger | Lewis | System Administrator | 32 | 3| 100000 | 13000 | roger@mail.com |
| 14 | Danny | Gibson | System Administrator | 31 | 2| 90000 | 12000 | danny@hotmail.com |
| 15 | Mike | Harper | Senior Marketing Executive | 36 | 1| 120000 | 28000 | m@gmail.com |
| 16 | Mary | Sunday | Marketing Executive | 31 | 5| 90000 | 25000 | monica@bigmail.com |
| 17 | Jack | Sim | Marketing Executive | 27 | 1| 70000 | 18000 | hal@gmail.com |
| 18 | Joe | Irvine | Marketing Executive | 27 | 1| 72000 | 18000 | joseph@hotmail.com |
| 19 | Henry | Ali | Customer Service Manager | 32 | 3| 70000 | 9000 | shahida@hotmail.com |
| 20 | Peter | Champion | Finance Manager | 32 | 2| 120000 | 25000 | peter@yahoo.com |
+----+-----------+----------+----------------------------+------+---------------+--------+-------+---------------------+
20 rows in set (0.01 sec)
mysql> select COUNT(*) from employee
-> where title = 'Programmer';
+----------+
| COUNT(*) |
+----------+
| 4 |
+----------+
1 row in set (0.00 sec)
*/
Drop table employee;
CREATE TABLE employee (
id int unsigned not null auto_increment primary key,
firstname varchar(20),
lastname varchar(20),
title varchar(30),
age int,
yearofservice int,
salary int,
perks int,
email varchar(60)
);
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("John", "Chen", "Senior Programmer", 31, 3, 120000, 25000, "j@hotmail.com");
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("Jan", "Pillai", "Senior Programmer", 32, 4, 110000, 20000, "g@yahoo.com");
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("Ane", "Pandit", "Web Designer", 24, 3, 90000, 15000, "a@gmail.com");
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("Mary", "Anchor", "Web Designer", 27, 2, 85000, 15000, "m@mail.com");
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("Fred", "King", "Programmer", 32, 3, 75000, 15000, "f@net.com");
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("John", "Mac", "Programmer", 32, 4, 80000, 16000, "j@hotmail.com");
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("Arthur", "Sam", "Programmer", 28, 2, 75000, 14000, "e@yahoo.com");
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("Alok", "Nanda", "Programmer", 32, 3, 70000, 10000, "a@yahoo.com");
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("Susan", "Ra", "Multimedia Programmer", 32, 4, 90000, 15000, "h@gmail.com");
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("Paul", "Simon", "Multimedia Programmer", 23, 1, 85000, 12000, "ps@gmail.com");
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("Edward", "Parhar", "Multimedia Programmer", 30, 2, 75000, 15000, "a@hotmail.com");
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("Kim", "Hunter", "Senior Web Designer", 32, 4, 110000, 20000, "kim@coolmail.com");
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("Roger", "Lewis", "System Administrator", 32, 3, 100000, 13000, "roger@mail.com");
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("Danny", "Gibson", "System Administrator", 31, 2, 90000, 12000, "danny@hotmail.com");
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("Mike", "Harper", "Senior Marketing Executive", 36, 1, 120000, 28000, "m@gmail.com");
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("Mary", "Sunday", "Marketing Executive", 31, 5, 90000, 25000, "monica@bigmail.com");
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("Jack", "Sim", "Marketing Executive", 27, 1, 70000, 18000, "hal@gmail.com");
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("Joe", "Irvine", "Marketing Executive", 27, 1, 72000, 18000, "joseph@hotmail.com");
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("Henry", "Ali", "Customer Service Manager", 32, 3, 70000, 9000, "shahida@hotmail.com");
INSERT INTO employee (firstname, lastName, title, age, yearofservice, salary, perks, email) values ("Peter", "Champion", "Finance Manager", 32, 2, 120000, 25000, "peter@yahoo.com");
select * from employee;
select COUNT(*) from employee
where title = 'Programmer';