php 根据日期显示星座的简单示例

想知道你是哪个星座的吗，这段代码只需要你输入生日，即可显示所在星座，php根据日期显示所在星座，感兴趣的小伙伴，下面一起跟随编程之家 jb51.cc的小编来看看吧。
经测试代码如下：

/**
 * 根据日期显示星座
 *
 * @param 
 * @author 编程之家 jb51.cc jb51.cc
 **/
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 



/***   代码来自编程之家 jb51.cc（jb51.cc）   ***/