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。