opendir php error with variables

Question

Can someone tell me what is wrong with this code?

<?php
    $user = $fgmembersite->UserFullName();
    $handle = opendir('/users/$user/');
?>

This is the error message:

opendir(/app/$user/) [function.opendir]: failed to open dir: No such file or directory in

score 3 · Accepted Answer · edited May 23 '17 at 12:11

3

Double quotes to evaluate variables

$handle = opendir("/users/$user/");

edited May 23 '17 at 12:11

Community

answered Mar 23 '12 at 17:09

Mike B

1

"doesn't work" isn't in a developer's vocabulary. You'll have to be more specific. – Mike B Mar 23 '12 at 17:11
then the directory doesn't exist, OR if it does then you are probably looking at a permissions issue – Murray McDonald Mar 23 '12 at 17:13
1

`/users` is going to be relative to the root. What you probably want is `opendir("users/$user/")` – wanovak Mar 23 '12 at 17:13
nevermind I figured it out and your solution helped me a lot wanovak. Thanks :) – Flame Trap Mar 23 '12 at 17:16
yes the directory didn't exist because I had a "/" which pointed it to the root directory of my server not the directory I was working in – Flame Trap Mar 23 '12 at 17:21

score 1 · Answer 2 · answered Mar 23 '12 at 17:11

1

In single quotes $user will not be interpreted as a variable - in double quotes it will be try opedir("/users/$user")

answered Mar 23 '12 at 17:11

Murray McDonald

score 1 · Answer 3 · answered Mar 23 '12 at 17:11

1

try this:

$handle = opendir("/users/$user/");

answered Mar 23 '12 at 17:11

magicianiam

3 Answers3