php 根据日期显示星座的解决办法

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/**
 * 根据日期显示星座
 *
 * @param 
 * @author php教程 www.idcnote.com
 **/
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php教程（www.idcnote.com）   ***/

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