Create table with birth days as follows birth_day field data type is date name the table as"person".
CREATE TABLE IF NOT EXISTS `person` (
`id` int(45) NOT NULL AUTO_INCREMENT,
`FirstName` varchar(15) DEFAULT NULL,
`LastName` varchar(15) DEFAULT NULL,
`birth_day` date DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;
PHP Codes to display age from birth is bellow.
<?php $host="localhost"; $username="root"; $password=""; $db_name="my_database"; $tbl_name="person"; mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $sql="SELECT id,FirstName,LastName,year(curdate())-year(birth_day) as Age FROM $tbl_name"; $result=mysql_query($sql); ?> <table align="center" bgcolor="#15475c"> <tr bgcolor="#ddeef5"> <td align="center" ><strong>ID</strong></td> <td align="center" "><strong>First Name</strong></td> <td align="center" "><strong>Last Name</strong></td> <td align="center" "><strong>Age</strong></td> </tr> <?php while($rows=mysql_fetch_array($result)){ ?> <tr bgcolor="#ddeef5"> <td bgcolor="#FFFFFF"><?php echo $rows['id']; ?></td> <td bgcolor="#FFFFFF"><?php echo $rows['FirstName']; ?></td> <td align="center" bgcolor="#FFFFFF"><?php echo $rows['LastName']; ?></td> <td align="center" bgcolor="#FFFFFF"><?php echo $rows['Age'];?></td> </tr> <?php } mysql_close(); ?> </table> |
You will see the final table like this.
ID | First Name | Last Name | Age |
1 | Scarl | Johan | 48 |
2 | Clooneer | Henderson | 38 |
3 | Katena | Daviddson | 24 |
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