php根据日期显示所在星座的方法

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我是靠谱客的博主甜美母鸡，最近开发中收集的这篇文章主要介绍php根据日期显示所在星座的方法，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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。