PHP Date Exercises : Calculate the current age of a person
PHP date: Exercise-18 with Solution
Write a PHP script to calculate the current age of a person.
Sample date of birth : 11.4.1987
<?php $bday = new DateTime('11.4.1987'); // Your date of birth $today = new Datetime(date('m.d.y')); $diff = $today->diff($bday); printf(' Your age : %d years, %d month, %d days', $diff->y, $diff->m, $diff->d); printf("\n"); ?>
Your age : 30 years, 3 month, 0 days
N.B.: The result may varry for your system date and time.
PHP Code Editor:
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PHP: Tips of the Day
SQL injection that gets around mysql_real_escape_string()
Consider the following query:
$iId = mysql_real_escape_string("1 OR 1=1"); $sSql = "SELECT * FROM table WHERE id = $iId";
mysql_real_escape_string() will not protect you against this. The fact that you use single quotes (' ') around your variables inside your query is what protects you against this. The following is also an option:
$iId = (int)"1 OR 1=1"; $sSql = "SELECT * FROM table WHERE id = $iId";
Ref : https://bit.ly/32q3bJ7
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