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php实现随机显示图片方法汇总

   本文分享一个php实现的随机显示图片的函数,可以将指定文件夹中存放的图片随机地显示出来。有兴趣的朋友研究下吧。

  php通过rand()函数产生随机数,这个函数可以产生一个指定范围的数字

  这段代码通过产生的随机数,随机选择图片

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srand( microtime() * 1000000 );

$num = rand( 1, 4 );

 

switch( $num )

{

case 1: $image_file = "/home/images/alfa.jpg";

break;

case 2: $image_file = "/home/images/ferrari.jpg";

break;

case 3: $image_file = "/home/images/jaguar.jpg";

break;

case 4: $image_file = "/home/images/porsche.jpg";

break;

}

echo "Random Image : php实现随机显示图片方法汇总";

?>

  方法二:

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$handle = opendir('./'); //当前目录

while (false !== ($file = readdir($handle))) { //遍历该php教程文件所在目录

list($filesname,$kzm)=explode(".",$file);//获取扩展名

if ($kzm=="gif" or $kzm=="jpg") { //文件过滤

if (!is_dir('./'.$file)) { //文件夹过滤

$array[]=$file;//把符合条件的文件名存入数组

}

}

}

$suiji=array_rand($array); //使用array_rand函数从数组中随机抽出一个单元

?>

php实现随机显示图片方法汇总

  方法三:

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/**********************************************

* Filename : img.php

* Author : freemouse

* Usage:

* php实现随机显示图片方法汇总

* php实现随机显示图片方法汇总

***********************************************/

if($_GET['folder']){

$folder=$_GET['folder'];

}else{

$folder='/images/';

}

//存放图片文件的位置

$path = $_SERVER['DOCUMENT_ROOT']."/".$folder;

$files=array();

if ($handle=opendir("$path")) {

while(false !== ($file = readdir($handle))) {

if ($file != "." && $file != "..") {

if(substr($file,-3)=='gif' || substr($file,-3)=='jpg') $files[count($files)] = $file;

}

}

}

closedir($handle);

$random=rand(0,count($files)-1);

if(substr($files[$random],-3)=='gif') header("Content-type: image/gif");

elseif(substr($files[$random],-3)=='jpg') header("Content-type: image/jpeg");

readfile("$path/$files[$random]");

?>

  以上所述就是本文的全部内容了,希望大家能够喜欢。

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