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Pulling MySQL for calculation using PHP

Posted on 2004-03-22
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Medium Priority
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Last Modified: 2013-12-12
This is the PHP that I have so far to display firstname, lastname, birthdate and email for a MySQL database that I have. I want to know now how to make the calculation for age, I have the SQL ready (see $age_mysql) but do not know how to send it to the database to get the calculated age in return displayed in the HTML page.

Also, is there an easy way to get the dates displayed other than in MySQL format YYYY-MM-DD?


This is my program to-date:

mysql_select_db($dbname, $DBConn);
$query_mysql = "SELECT * FROM birthdays";
$mysql = mysql_query($query_mysql , $DBConn) or die(mysql_error());
$row_mysql = mysql_fetch_assoc($mysql );
$totalRows_mysql = mysql_num_rows($mysql );
$age_mysql = "(YEAR(CURDATE())-YEAR(birthdate)) - (RIGHT(CURDATE(),5)<RIGHT(birthdate,5))";
?>

<html>
<head>
<title>Birthdays</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<body>
<table width=75%>
<tr>
<td width=20%>Last Name</td>
<td width=20%>First Name</td>
<td width=20%>Date Of Birth</td>
<td width=20%>Email</td>
<td width=20%>Age</td>
</tr>
<?php do { //Begin the loop to iterate through the records ?>
<tr>
<td><?php echo $row_mysql['lastname']; ?></td>
<td><?php echo $row_mysql['firstname']; ?></td>
<td><?php echo $row_mysql['birthdate']; ?></td>
<td><?php echo $row_mysql['email']; ?></td>
<td>the age calculation to be inserted here</td>
</tr>
<? } while ($row_mysql = mysql_fetch_assoc($mysql )); //End the loop ?>
</table>
</body>
<?php
mysql_free_result($mysql);
//And close the connection
mysql_close($DBConn);
?>
0
Comment
Question by:Rusty20009
  • 2
3 Comments
 
LVL 6

Assisted Solution

by:CosminB
CosminB earned 100 total points
ID: 10648610
you could use the date_format function from mysql
or you could format it using php, using something like this:
<?
$dateString = '2004-03-21';
echo date('r', strtotime($dateString));
?>

so you birthdate line should look like this:

<td><?php echo date('f D Y', strtotime($row_mysql['birthdate'])); ?></td>

look at http://www.php.net/date for more info on date()
0
 
LVL 8

Accepted Solution

by:
william_jwd earned 300 total points
ID: 10648727
try this,

mysql_select_db($dbname, $DBConn);
$query_mysql = "SELECT lastname, firstname, birthdate, email, (YEAR(CURDATE())-YEAR(birthdate)) - (RIGHT(CURDATE(),5)<RIGHT(birthdate,5)) as age FROM birthdays";
$mysql = mysql_query($query_mysql , $DBConn) or die(mysql_error());
$row_mysql = mysql_fetch_assoc($mysql );
$totalRows_mysql = mysql_num_rows($mysql );
?>

<html>
<head>
<title>Birthdays</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<body>
<table width=75%>
<tr>
<td width=20%>Last Name</td>
<td width=20%>First Name</td>
<td width=20%>Date Of Birth</td>
<td width=20%>Email</td>
<td width=20%>Age</td>
</tr>
<?php do { //Begin the loop to iterate through the records ?>
<tr>
<td><?php echo $row_mysql['lastname']; ?></td>
<td><?php echo $row_mysql['firstname']; ?></td>
<td><?php echo $row_mysql['birthdate']; ?></td>
<td><?php echo $row_mysql['email']; ?></td>
<td><?php echo $row_mysql['age']; ?></td>
</tr>
<? } while ($row_mysql = mysql_fetch_assoc($mysql )); //End the loop ?>
</table>
</body>
<?php
mysql_free_result($mysql);
//And close the connection
mysql_close($DBConn);
?>
0
 
LVL 6

Expert Comment

by:CosminB
ID: 10648902
and if you want to find the age try this:
<?php
$dateString = '1980-03-21';
$age = date('Y', time()-strtotime($dateString))-1970;
echo $age;
?>
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