Multipart form $_FILES array is empty after submission, but files upload

Question

I have a multipart form in which I allow the user to select multiple files which is placed in an array. Here is a scaled down portion of my code:

<form id="fileupload" action="upload.php" method="POST" enctype="multipart/form-data">
    <input type="file" name="files[]" multiple>
</form>

In my upload.php file, I want to write to a text file and place info from each file (doesn't matter how it's formatted, array is fine).

$date = $_POST["date"];
$files = $_POST["files"];


$myFile = $date . ".txt";
$fh = fopen($myFile, 'w') or die("cant_open_file");
$stringData = "Date: " . $date . "\n";
$stringData .= "Uploaded Files:\n";
$stringData .=  $file; // prints nothing?
fwrite($fh, $stringData);
fclose($fh);

However, nothing is $file is printing nothing.

Question

How can I print the file info for each file that is chosen from the user into my text file?

Desired output

Date: Todays Date
Files: 2 file(s)
    File: someimage1.png
    Size: 10KB

    File: someimage1.png
    Size: 20KB

Update

For some reason, $_FILES['files'] is blank when submitted but I when I use the preceding example, "File: " and "Size: " print out, but $file['name'] and $file['size'] seem to be blank. Also, say I choose 3 files, "File/Size" is printed 5 times (suggesting that 5 files were chosen). This is the same for any amount of files chosen. I've made sure the correct values are in php.ini as well. Not sure where to go from here.

foreach($files as $file){
    $stringData .=  "   File: ".$file['name']."\n"; // return blank
    $stringData .=  "   Size: ".$file['size']."\n"; // return blank
}

Also, I am using jQuery File Upload Demo.

xdazz · Accepted Answer · 2012-09-06 06:49:25Z

4

Use $files = $_FILES["files"]; to get the uploaded file information.

POST method uploads

answered Sep 6, 2012 at 6:49

xdazz

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Comments

Mihai Iorga · Accepted Answer · 2012-09-06 06:50:26Z

1

You can use loop in the $_FILES array:

foreach($files as $file){
    $stringData .=  "File: ".$file['name']."\n";
    $stringData .=  "Size: ".$file['size']."\n";
}

make sure you use move_uploaded_file() because you only store them in /tmp folder and you will lose them.

answered Sep 6, 2012 at 6:50

Mihai Iorga

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2 Comments

WrightsCS Over a year ago

Okay, so a little progress made. Now i get "File: " and "Size: " but $file['name'] and $file['size'] seem to be blank. Also, say I choose 3 files, "File/Size" is printed 5 times (suggesting that 5 files were chosen. Not sure what thats all about.

WrightsCS Over a year ago

Actually, it' doesn't matter how many files I select, it always shows as 5 files. Any idea why that would be?

Abid Hussain · Accepted Answer · 2012-09-06 06:50:42Z

1

Use

$files = $_FILES['files'];

echo "<pre>";
print_r($files);

instead of

$files = $_POST["files"];

answered Sep 6, 2012 at 6:50

Abid Hussain

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Comments

Prabhuram · Accepted Answer · 2012-09-06 06:51:01Z

1

Please try using $_FILES as follows instead of $_POST['files']:

print_r($_FILES['files'])

answered Sep 6, 2012 at 6:51

Prabhuram

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