displaying of files in a directory in a select menu

[william232]

Hi all,

I am trying to create a snippet of developing a viewing files in a select/list menu being dynamically created so it displays all files in a certain directory

only problem is that when i do this i get this back for some reason

<br/> contents of the directory:rank<br/> <select name="cnt">
<option value="">Please a file</option><option value=1">1</option><option value="">Please a file</option><option value=1">1</option><option value="">Please a file</option><option value=1">1</option> </select>


now why am i getting 1 for value and name of the value??

can anyone help me out here please,

Here is the code i am using

    $dir="rank";
    $odir=opendir($dir);//will open the directory
    echo "<br/> contents of the directory:".$dir."<br/>";
    ?>
    <select name="cnt">
    <?
    while($contents=readdir($odir))//read the contents of the directory
    {
        if($contents="." || $contents="..")
        {
            echo "<option value=\"\">Choose a file</option>";    
            echo "<option value=".$contents."\">".$contents."</option>";
        }
        
    }?>
    </select>
    <?
    closedir($odir);//closes the directory that is open

Thanks,William

What am i doing wrong???

[spikeZ]

personally I would use glob() rather than opendir.

<?php
$dir = '.';
$option = '';
$files = glob($dir.'/*');
foreach($files as $key=>$data) {
    $option .= '<option>'. $data .'</option>';
}

?>
<select name="test">
    <?php echo $option; ?>
</select>

[Salathe]

The quick answer is that you're using = where you should be using == in your if statement.

Off Topic:

I did have a long, rambling answer prepared as to why the value of $contents echoes as 1 but I closed the browser tab and lost it all.