How do you find images with a name similar to a variable? - php

Question

So far, I wrote a script which finds if a certain file exists in an external website.
<?php
if (isset($_GET['user'])){
if ($_GET['user'] != ""){
$userloc = "http://somesite.com/images/" . htmlspecialchars($_GET["user"]) . ".png";
$user = htmlspecialchars($_GET['user']);
if (getimagesize($userloc) !== false) {
echo $user . "exists!" ;
echo "<img src=\"" . $userloc . "\" height=\"100px\" />";
} else {
echo $username . "\" does not exist.";
$suggest = /* Code to find an image with a similar name */
echo "Did you mean: ";
}
}
?>
Basically, I would like to expand the code so it will display image suggestions.
I am a complete newbie when it comes to PHP code. Any help would be much appreciated. :)

Answer 1

The levenshtein function is useful for finding similarities in strings.
Assuming that you have correctly checked to see that a file exists or not, then you can use this function to make suggestions.
/**
* #param string $inputName
* #param string[] $knownNames
*
* #return string[]
*/
function getNameSuggestions($inputName, array $knownNames = array())
{
$candidates = array();
foreach ($knownNames as $candidate) {
$lev = levenshtein($inputName, $candidate);
if ($lev <= strlen($inputName) / 3 || false !== strpos($candidate, $inputName)) {
$candidates[] = $candidate;
}
}
return $candidates;
}
Then call the function
$suggestions = getNameSuggestions((string) $_GET['user'], array(
// a list of known users
));

Answer 2

Your script does not test if a file exist on a remote website. What it does it assumes that the filename $_GET['user'] . 'png'; exists at http://somesite.com/images/ if it's not ""
What you should do is something like this:
$remote = 'http://somesite.com/images/' . urlencode($_GET["user"]) . '.png';
if (file_get_contents($remote) !== false) {
echo 'Exists!';
} else {
echo 'Does not exist!';
// try something else or start bruteforce :P
}