php如何根据日期显示所在星座？
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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