Showing Images in PHP

Question

I want to show images residing inside a directory in PHP The code which I wrote was

<?php

$dir = opendir("upload");


while (($file = readdir($dir)) !== false)
  {
  echo "filename: " . $file . "<br />";

  echo "<a href='shoe.php?image_name=".$file."'> <img src='upload/".$file[0]."' width='150' height='150' /></a></td>";

  }
closedir($dir);
?>

Here the images are not displayed. Only a frame comes for that image and the actual image is not displayed inside that. Where I have went wrong? Can any one please help...

score 1 · Answer 1 · answered Apr 02 '11 at 20:34

1

PHP Glob makes it much easier to list directories. You can do something like

$image = glob("*.jpg")
foreach ($image as $filename) {
    echo "$filename size " . filesize($filename) . "\n";
}

answered Apr 02 '11 at 20:34

amosrivera

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amosrivera: Thanks amosrivera – SK0004 Apr 02 '11 at 20:37

score 0 · Accepted Answer · answered Apr 02 '11 at 20:33

0

Replace $file[0] with $file, as $file is not an array and that will only return the first character of the string.

answered Apr 02 '11 at 20:33

score 0 · Answer 3 · answered Apr 02 '11 at 20:37

why $file[0]??? $file is correct:

<?php

        $dir = opendir("upload");

        while (($file = readdir($dir)) !== false)
          {
          echo "filename: " . $file . "<br />";

          echo "<a href='shoe.php?image_name=".$file."'> <img src='upload/".$file."' width='150' height='150' /></a>";

          }
        closedir($dir);
    ?>

Showing Images in PHP

3 Answers3

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