将$ _Files ['userfile'] ['name']转换为变量

Question

I am using code for an to upload files as seen below:

$uploaddir = "./images/";
$uploadfile = $uploaddir . basename($_FILES['userfile']['name']);

if (move_uploaded_file($_FILES['userfile']['tmp_name'], $uploadfile)) {
  echo "File is valid, and was successfully uploaded.
";
} else {
   echo "Upload failed";
}

The code works to upload the image but I want to be able to post the filename to an SQL database afterwords.

I need to get the filename I uploaded and make it a variable called $filename so i can then use:

$image = $_POST[$filename];

I've tried using this to make it a variable

$filename = $_FILES['userfile']['name'];

All I get is an error saying undefined index : image.jpg on the $_POST and I'm uncertain why. Its listing the image filename in the error so why cant it upload it.

Answer 1

You can only access $_POST[$filename] if you're posting an input with the name "image.jpg", which I guess you aren't.

You've already got the filename from the $_FILES array, so why not use it again?

$image = basename($_FILES['userfile']['name']);

... or use the $filename variable that you've already got (can't see where this is defined in your example, but it obviously contains "image.jpg")

Note: the name part of $_FILES['input_name'] arrays contains the original filename of the uploaded file. That's all you need.