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I am trying to populate an HTML form's drop down list with the contents of the current directory. I attempted to use PHP's scandir() function to output these files as an array and then feed them into the option's value tag within the HTML form.

I looked at various solutions available on SO and outside, but none seemed to work for me.

This is how my code looks like right now:

<form action="" >
        <input type="submit" class="Links" value="Select Input">
        <select >               
            <?php 
                $dir="./inputfiles/"; 
                $filenames = scandir($dir); 
                foreach($filenames as $file){
                    echo '<option value="' . $filenames . '">' . $filenames . '</option>';
                    }
            ?>                      
        </select> 
</form>

For now I'm getting absolutely no options in the drop down menu.

I am very new to PHP and would appreciate your feedback as to how to make this work.

Further Changes:

I tried the solutions given below, and none seemed to work. I changed the code to check whether the PHP script is able to output any variable or not in the html forms list.

<form action="" >
        <input type="submit" class="Links" value="Select Input">
        <select >               
            <?php 
                  $someval = "Video";
                  echo '<option value="value"> ' . $someval .' </option>';
            ?>                      
        </select> 
</form>

It displays ' . $someval .' instead of Video in the menu bar

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3
  • Also, make sure to use 2 periods to go up a level in a file system: $dir="../inputfiles/"; Commented Dec 14, 2012 at 0:17
  • I am searching for files in the current level onwards. So, inputfiles folder is present in the current directory. Commented Dec 14, 2012 at 0:21
  • 1
    @NineHourz In that case, you should use the path inputfiles/ Commented Dec 14, 2012 at 0:25

3 Answers 3

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2

You have to use $file in the echo out not $filenames. Like this:

echo '<option value="' . $file . '">' . $file . '</option>';

See PHP foreach for more info.

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3 Comments

Thank you for your quick reply. I tried this change, but it still gives no change in the output.
Check the path. Or may be try with absolute path first to see if it work. You can also use function getcwd (php.net/manual/en/function.getcwd.php) to get the current working path to confirm that you get it right.
That is kind of weird. Just for the sake of trying: can you echo each part for each line (so there will be 5 lines).
2

When iterating over the array, you need to use $file for concatenation, not $filenames:

echo '<option value="' . $file . '">' . $file . '</option>';

If inputfiles is present in the same directory, you should use the path: inputfiles/

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Comments

1

A couple of concerns with your script.

First, you need to output the correct variable within the options:

echo '<option value="' . $file . '">' . $file . '</option>';

Second, you will likely need to add detection for . and .. in your loop, as you will likely not want to output them. You might also want to exclude directories as well.

Third, you should probably make it clear exactly what directory you are trying to scan. Try using a full file path. If you need it relative to the currently executing file's directory you can do something like:

$dir= __DIR__ . DIRECTORY_SEPARATOR . 'inputfiles'; // no need for trailing slash

Fourth, you might want to consider working with something like DirectoryIterator to give you some more convenient methods to do what you are looking to do (i.e. verifying it's an actual file before listing it). So something like this:

$dir= __DIR__ . DIRECTORY_SEPARATOR . 'inputfiles';
$iterator = new DirectoryIterator($dir);
foreach ($iterator as $fileinfo) {
    if($fileinfo->isFile()) {
        echo '<option value="' . $fileinfo->getFilename() . '">' . $fileinfo->getFilename() . '</option>';
    }
}

In this trivial example, it you won't necessarily gain a lot (in terms of saving extra code) by using DirectoryIterator, but I certainly like to point out it's usage when I can, because if you needed to start doing things like showing file size, file permissions, file modification times, etc. It is much easier to do it using the DirectoryIterator class (or RecursiveDirectoryIterator class if you need to recursive) than using traditional PHP's file system functions.

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4 Comments

Thank you very much Mike! Your explanation makes sense. I tried out your suggestion but it still shows ' . $fileinfo->getFilename() . ' in the menu bar instead of the actual variable value. If I replace the line with echo '<option value="' . $fileinfo->getFilename() . '">somevalue</option>'; it does show somevalue in the menu bar. I don't understand why php isnt able to parse the actual value correctly.
Do the $fileinfo object exists otherwise the isFile() would not equal true. Are you sure you have the quotes in your echo placed correctly? What do you get when you var_dump($fileinfo) or var_dump($fileinfo->getFilename())?
When I print out the values from var_dump($fileinfo) onto a separate browser, I get the correct list of files. However, when I include var_dump.. within the value option, it prints as it is without evaluating to the actual value.
@NineHourz You must be having some sort of unclosed quotes, unbalanced <?php and ?> blocks, or something of that nature. When you look at the HTML source for your page, do you see PHP content throughout? If you only see certain portions of PHP, look at where it starts to show up and that is likely where you are missing a <?php opening.

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