从目录中拉出图像 - PHP

问题描述：

I am trying to pull images simply from my directory /img and load them dynamically into the website into the following fashion.

            <img src="plates/photo1.jpg">

That's it. It seems so simple but all of the code I have found basically doesn't work.

What I have that I am trying to make work is this:

   <?php
   $a=array();
   if ($handle = opendir('plates')) {
while (false !== ($file = readdir($handle))) {
   if(preg_match("/\.png$/", $file)) 
        $a[]=$file;
else if(preg_match("/\.jpg$/", $file)) 
        $a[]=$file;
else if(preg_match("/\.jpeg$/", $file)) 
        $a[]=$file;

}
closedir($handle);
   }

 foreach($a as $i){
echo "<img src='".$i."' />";
 }
 ?>

我正在尝试从我的目录/ img中提取图像，并将它们动态加载到网站中，形式如下。 p>

 ＆lt; img src =“plates / photo1.jpg”＆gt; 
  code>  pre> 
 
 就是这样。 它似乎很简单，但我发现的所有代码基本上都不起作用。 p> 
 
 
我想要做的工作是： p> 
 
  
 ＆lt;？php 
 $ a = array（）; 
 if（$ handle = opendir（'plates'））{
while（false！==（$ file = readdir（$ handle）  ））{
 if（preg_match（“/ \。png $ /”，$ file））
 $ a [] = $ file; 
else if（preg_match（“/ \。jpg $ /”，$ file）  ）
 $ a [] = $ file; 
else if（preg_match（“/ \ .jpeg $ /”，$ file））
 $ a [] = $ file; 
 
} 
 nclosedir（$ handle  ）; 
} 
 
 foreach（$ a as $ i）{
echo“＆lt; img src ='”。$ i。“'/＆gt;”; 
} 
？＆gt; 
   pre> 
  div>

答

You want your source to show up as plates/photo1.jpg, but when you do echo "<img src='".$i."' />"; you are only writing the file name. Try changing it to this:

<?php
$a = array();
$dir = 'plates';
if ($handle = opendir($dir)) {
  while (false !== ($file = readdir($handle))) {
    if (preg_match("/\.png$/", $file)) $a[] = $file;
    elseif (preg_match("/\.jpg$/", $file)) $a[] = $file;
    elseif (preg_match("/\.jpeg$/", $file)) $a[] = $file;
  }
  closedir($handle);
}
foreach ($a as $i) {
  echo "<img src='" . $dir . '/' . $i . "' />";
}
?>

答

You should use Glob instead of opendir/closedir. It's much simpler.

I'm not exactly sure what you're trying to do, but you this might get you on the right track

<?php
foreach (glob("/plates/*") as $filename) {

    $path_parts = pathinfo($filename);

    if($path_parts['extension'] == "png") {
        // do something
    } elseif($path_parts['extension'] == "jpg") {
        // do something else
    }
}
?>

答

This can be done very easily using glob().

$files = glob("plates/*.{png,jpg,jpeg}", GLOB_BRACE);
foreach ($files as $file)
    print "<img src=\"plates/$file\" />";

从目录中拉出图像 - PHP

相关推荐