[RESOLVED] Populating a Drop Down via a file directory?

suttercain

Hi everyone,

I just wrote this code:

$dir = "/images/news";
print "<select name='file'>";
$dir = opendir($dir);
while (false !== ($file = readdir($dir))){
if (in_array($file, array(".", ".."))) continue;
print "<option value='$file'>$file</option>";
}
print "</select>";

But for some reason it is not populating. It instead echoes an empty drop down menu. Can anyone suggest anything?

Thank you for your help.

scross

Your code seems fairly sound, I'm just not sure you mean't to include the '/' at the beginning of your directory name, since this refers to the root of the server from an internal point of view and so php searches for a directory called "images" in the root of your server. I apologise if your directory is correct, however when I changed the directory to one which I knew existed it worked perfectly, so this seems the most likely error.

suttercain

Thank you. That worked.

I am now having a problem with the selection result. The above code is within a form. The user selects the image and presses submit.

When they are taken to the next page the image is echoed. Only thing is, the same image is echoed over and over no matter what selection is made. If image1.jpg is select it shows image1.jpg, if image2.jpg is selected it still shows image1.jpg

Anyone know why?

Thanks again for the help.

scross

I've tested it and the above code seems perfect sound, you'll need to provide some more code so I can locate where the error might be.

suttercain

Okay I got it populated via the file directory. I had a slash before images when I shouldn't have.

BUT...Now I am having another issue with it. The code is in a form. The user selects the image from the drop down and submits it. One the next page the image they selected is echoed in the browser.

PROBLEM:No matter which image they select the same one comes up over and over.

Here is the Form Code:

<form action="preview_news.php" method="post">
<select name='image'>
			<?php
$dir = "images/news";
$dir = opendir($dir);
while (false !== ($image = readdir($dir))){
if (in_array($image, array(".", ".."))) continue;
?>
<option value='<?php echo $image; ?>'><?php echo $image; ?></option>
<?php
}
?>
</select>

And Here is the page that echos the picture (should be) based on the selection:

<?php
session_start();
$_SESSION['image'] = $_POST['image'];
$image = $_SESSION['image'];
echo "<b>IMAGE:</b><br><img src='images/news/" . $row['image'] ."' /><br><br>";
?>

I am at a loss. I got the menu populated... but it's ignoring my selection...

Any ideas?

Thanks guys!

scross

Where does "$row['image']" come from in:

echo "<b>IMAGE:</b><br><img src='images/news/" . $row['image'] ."' /><br><br>";

...since you set the variable $image to the selected image but then you output "$row['image']", it looks like this could be the source of your error...

suttercain

Doh! I am lame. Thank you. I was still getting it from my database when I should just get it from the directory... $image.

Thank you for your time. I really appreciate it.

scross

No problem at all 🙂