Some companies will be given trainings to their staff members.In this case they maintain data base with 2 tables one with staff members details and other with trainings details.when all staff members are not trained higher executives need to know the staff members who have not trained yet.Here with this problem left joining of 2 tables in following manner executives could be easily found the staff members have not trained yet.
I have used bellow 2 tables for this purpose.
According to these two tables you can see jennifer,sofia & wictor have not got any training. dept_division table "id" is used to identify the staff members in table training as "training_id".
Create following php file and run it and you get the expected result.
<?php $host="localhost"; $username="root"; $password=""; $db_name="my_database"; mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $sql ="SELECT dept_division.name, dept_division.age,dept_division.division FROM dept_division LEFT JOIN training ON training.traning_id=dept_division.id WHERE training.traning_id IS NULL"; $result=mysql_query($sql); ?> <table align="center" bgcolor="#15475c"> <tr bgcolor="#ddeef5"> <td align="center" ><strong>Name</strong></td> <td align="center" "><strong>Age</strong></td> <td align="center" "><strong>Division</strong></td> </tr> <?php while($rows=mysql_fetch_array($result)){ ?> <tr bgcolor="#d0e7f1"> <td><?php echo $rows['name']; ?></td> <td><?php echo $rows['age']; ?></td> <td><?php echo $rows['division']; ?></td> </tr> <?php } ?> </table> |
Resulting table is
Name | Age | Division |
jennifer | 23 | Production |
sofia | 25 | service |
wictor | 45 | production |
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