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。
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