I have a php code that has a dropdown menu filled with the names of images files in a specific folder, what i want is once the user select one of the names the selected image is displayed, at the moment i'm trying to use a submit button. i am trying to not use jquery or other languages
<?php
$folder = 'C:\Users\source\\';
echo '<form action="" method="post">'."\n".'<select name="image">'."\n".
dropdown(image_filenames($folder), @$_POST['image']).
'</select>'."\n".'<input type="submit" name="submit">'."\n".'
</form>';
function image_filenames($dir)
{
$handle = @opendir($dir)
or die("I cannot open the directory '<b>$dir</b>' for reading.");
$images = array();
while (false !== ($file = readdir($handle)))
{
if (preg_match('/^.*\.(jpg|jpeg|png|gif|svg)$/', $file))
{
$images[] = $file;
}
}
closedir($handle);
return $images;
}
function dropdown($options_array, $selected = null)
{
$return = null;
foreach($options_array as $option)
{
$return .= '<option value="'.$option.'"'.
(($option == $selected) ? ' selected="selected"' : null ).
'>'.$option.'</option>'."\n";
}
return $return;
}
if (isset($_POST['submit'])) {
echo '<img src=C:\Users\source'. $row['images'] . ' />';
}
?>
You must need to keep your file inside your project ...
change
to
images is the folder name which inside your project...
xampp or wampp not consider the local path files...
And it won't display the images as well...
Hope this Helpful...
I have changed two places in your code...