Make a PHP variable be recognized [duplicate] - php

Question

This question already has answers here:
What is the difference between single-quoted and double-quoted strings in PHP?
(7 answers)
Closed 8 years ago.
I've been working on a php dynamic drop down list that scans a folder on the web server and list its contents it works hears relevant selection of the code.
<?php
foreach(glob(dirname(rootdir) . '/path/username/*') as $filename){
$filename = basename($filename);
echo "<option value='filepath/username/" . $filename . "'>".$filename."</option>";
}
?>
However if i use a variable to populate part of the file path the variable doesn't get added to the path.
<?php
$name = "jsmith";
foreach(glob(dirname(rootdir) . '/path/$name/*') as $filename){
$filename = basename($filename);
echo "<option value='filepath/$name/" . $filename . "'>".$filename."</option>";
}
?>
$name = "jsmith"
The result I'm looking for is /path/$name/* = /path/jsmith/
and
filepath/$name/ = filepath/jsmith/
How do I get the glob(dirname) and option value=' recognize the variables?

Answer 1

You need to either concatenate the variable with the string or wrap the string in double quotes so the variable is parsed.
Double Quotes
<?php
$name = "jsmith";
foreach(glob(dirname(rootdir) . "/path/$name/*") as $filename){
$filename = basename($filename);
echo "<option value='filepath/$name/" . $filename . "'>".$filename."</option>";
}
?>
Concatenate
<?php
$name = "jsmith";
foreach(glob(dirname(rootdir) . '/path/' . $name . '/*') as $filename){
$filename = basename($filename);
echo "<option value='filepath/$name/" . $filename . "'>".$filename."</option>";
}
?>

Answer 2

Single quotes do not interpolate. (i.e. Variables in the string will not be replaced with their values.) However, double-quotes do interpolate.
You'll need to change '/path/$name/*' to "/path/$name/*" or use string concatenation with '/path/' . $name . '/*' instead.
Also, you should be careful with accepting user-controlled input for $name as it may lead to a directory traversal attack.