Jochem -
The problem is the sort()! I did not realize that it replaces the key with an integer. Instead, I should have used ksort().
$dir = opendir($doc_dir);
while (false !== ($file = readdir($dir))) {
if (($file != ".") && ($file != ".."))
$file_list[$file] = $file;
}
// Debug code
echo '<pre>';
print_r($file_list);
echo '</pre>';sort($file_list); <-----------<<<<
foreach ($file_list as $key => $val) {
echo $key . " => " . $val . "<br>";
}Here are the result:
Array
(
[mailings] => mailings
[cases] => cases
)0 => cases 1 => mailings Todd
Jochem Maas wrote:
Todd Cary wrote:
I am using an array to populate a drop-down and I would like to have the same value in the key as the value. Using the following, the key is 0,1,2,3...etc. How can I correct that?
what is there to correct - you are telling me that the value of $file show up as a string and then magically it becomes an integer (but just while you are designating the key)
listen you've just shown us a simple loop to stuff some file/dir names into an array where by the file/dir name is the assoc. key as well as the value of any given item in said array - WHERE THE F*** IS THE DROPDOWN RELATED CODE?
$file_list = array(); $dir = opendir($doc_dir); while (false !== ($file = readdir($dir))) { if (($file != ".") && ($file != "..")) $file_list[$file] = $file; <---------------<<<<< }
what happens if you place the following right after the code you mention (just to prove that what you say about the key is wrong):
echo '<pre>'; print_r($file_list); echo '</pre>';
Todd
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