php根据日期显示所在星座的方法

本文实例讲述了php根据日期显示所在星座的方法。分享给大家供大家参考。具体实现方法如下：

<?php function zodiac($DOB){   $DOB = date("m-d", strtotime($DOB));   list($month,$day) = explode("-",$DOB);   if(($month == 3 || $month == 4) && ($day > 22 || $day < 21)){     $zodiac = "Aries";   }   elseif(($month == 4 || $month == 5) && ($day > 22 || $day < 22)){     $zodiac = "Taurus";   }   elseif(($month == 5 || $month == 6) && ($day > 23 || $day < 22)){     $zodiac = "Gemini";   }   elseif(($month == 6 || $month == 7) && ($day > 23 || $day < 23)){     $zodiac = "Cancer";   }   elseif(($month == 7 || $month == 8) && ($day > 24 || $day < 22)){     $zodiac = "Leo";   }   elseif(($month == 8 || $month == 9) && ($day > 23 || $day < 24)){     $zodiac = "Virgo";   }   elseif(($month == 9 || $month == 10) && ($day > 25 || $day < 24)){     $zodiac = "Libra";   }   elseif(($month == 10 || $month == 11) && ($day > 25 || $day < 23)){     $zodiac = "Scorpio";   }   elseif(($month == 11 || $month == 12) && ($day > 24 || $day < 23)){     $zodiac = "Sagittarius";   }   elseif(($month == 12 || $month == 1) && ($day > 24 || $day < 21)){     $zodiac = "Cpricorn";   }   elseif(($month == 1 || $month == 2) && ($day > 22 || $day < 20)){     $zodiac = "Aquarius";   }   elseif(($month == 2 || $month == 3) && ($day > 21 || $day < 21)){     $zodiac = "Pisces";   }   return $zodiac; } echo zodiac('1986-07-22'); //Valid strtotime date ?>

希望本文所述对大家的php程序设计有所帮助。