PHP Forum

**skima** · 05-27-2005, 07:42 PM

Please I want to display information from the database where date of birth has been stored. But i want to display the one with its day equals today's date.
This is the date formate i stored 0000-00-00. yy-mm-dd.

can anyone help out.

**rcubes85** · 05-28-2005, 02:33 AM

PHP Code:


			




 <?php ////// Check BirthDays: Error Code = 1

   $today = split("-", date("Y-m-d"));

  $result = mysql_query("SELECT Member, FirstName, Birthday FROM members;") or die ("Error Code #1.");

  while ($row = mysql_fetch_array($result)) {

      $birthday = split("-", $row[Birthday]);

      if (((int)$today[1] - (int)$birthday[1] == 0) && ((int)$today[2] - (int)$birthday[2] == 0)) {

        $age = (int)$today[0] - (int)$birthday[0];

        print "<tr bgcolor=\"#333333\" bordercolor=\"#333333\"><td colspan=\"5\" width=\"10%\">";

        print "<p>Hay what up <b>".$row[FirstName]."</b>, it's ya' birthday!!!<br>";

        print "who turns ".$age." today!<br></p></td></tr>";

        }

   }

  

  ?>



That might help

**skima** · 05-30-2005, 10:17 PM

Thanx for response. am trying it all out.

**skima** · 05-31-2005, 01:09 AM

Thank you rcubes85. It really works.

**rcubes85** · 05-31-2005, 01:47 AM

no problem glad it does