unable to echo multiple images with multiple variable - php

Question

I am uploading multiple images from a single input and rename all uploaded files. My problem is; I need to display all uploaded images with separate variables
For example:
If user upload 3 images, i need to echo as
$upload1 = 1st file name
$upload2 = 2nd file name
$upload3 = 3rd file anme
No of images uploaded (upload number of file)
Here is my current code:
<?php
if (isset($_FILES['files'])) {
$uploadedFiles = array();
foreach ($_FILES['files']['tmp_name'] as $key => $tmp_name) {
$errors = array();
$file_name = md5(uniqid("") . time());
$file_size = $_FILES['files']['size'][$key];
$file_tmp = $_FILES['files']['tmp_name'][$key];
$file_type = $_FILES['files']['type'][$key];
if($file_type == "image/gif"){
$sExt = ".gif";
} elseif($file_type == "image/jpeg" || $file_type == "image/pjpeg"){
$sExt = ".jpg";
} elseif($file_type == "image/png" || $file_type == "image/x-png"){
$sExt = ".png";
}
if (!in_array($sExt, array('.gif','.jpg','.png'))) {
$errors[] = "Image types alowed are (.gif, .jpg, .png) only!";
}
if ($file_size > 2097152000) {
$errors[] = 'File size must be less than 2 MB';
}
$desired_dir = "user_data/";
if (empty($errors)) {
if (is_dir($desired_dir) == false) {
mkdir("$desired_dir", 0700); // Create directory if it does not exist
}
if (move_uploaded_file($file_tmp, "$desired_dir/" . $file_name . $sExt)) {
$uploadedFiles[$key] = array($_FILES['files']['name'][$key], 1);
} else {
echo "Couldn't upload file " . $_FILES['files']['name'][$key];
$uploadedFiles[$key] = array($_FILES['files']['name'][$key], 0);
}
} else {
print_r($errors);
}
}
foreach ($uploadedFiles as $key => $row) {
if (!empty($row[1])) {
$codestr = '$file' . ($key+1) . " = $row[0];";
eval ($codestr);
} else {
$codestr = '$file' . ($key+1) . " = NULL;";
eval ($codestr);
}
}
}
echo $file1;
echo $file2;
echo $file3;
echo $file4;
echo $file5;
?>
I tried this and I get the error
Parse error: syntax error, unexpected 'png' (T_STRING) in C:\Users\logon\Documents\NetBeansProjects\upload file rename single and multiple\multiple file rename\method 2\process.php(43) : eval()'d code on line 1
What am I doing wrong? can some one help me

Answer 1

I think the " = $row[0];" on line 42 might not do what you want it to do. Because of the double quotes, the value of the variable $row[0] will be printed in the code.
The code running through the eval will look something like this:
$file0 = FILENAME.png;
Instead of this:
$file0 = $row[0];
To solve the problem you could simply turn line 42 into the following:
$codestr = '$file' . ($key+1) . ' = $row[0];';
More information about the difference between single and double quotes: http://php.net/manual/en/language.types.string.php
Edit: For the correct names of the files you could change $_FILES['files']['name'][$key] to $file_name . $sExt. Hope it helps!