Php function calling, Undefined function error

Question

Earlier I have a form that ask user to upload a picture and I have this function:

function fileUploaded() {
        $fileName = $_FILES ['picture'] ['name'];
        $pathOfFile = "/images/";
        $fileTmpLoc = $_FILES ['picture'] ["tmp_name"];
        $fileResult = move_uploaded_file ( $fileTmpLoc, $pathOfFile );
        if (isset ( $fileName )) {
            return true;
        }

}

Basically it moves the uploaded picture to images file. Then I am calling this function in an if statement:

        if (fileUploaded () == true) {
        if ($fileResult) {
          /*checking the size of file*/
            }
        }
       else {
        $fileName = "default.jpg";
       }

After when I try to upload and submit it gives the error in the below:

Fatal error: Call to undefined function fileUploaded()

What should be the problem? Thanks.

When you say 'I have this function', where is it written? In the same file, or a different file? If it's in a different file, are you definitely including it? — Dan Blows
– Dan Blows, Commented Jan 16, 2014 at 21:14
If you're calling the function "before" the function, then that will throw you that error. It depends on where it's placed. Show us the full code in order to see where it's placed, IF it's inside the same file. — Funk Forty Niner
– Funk Forty Niner, Commented Jan 16, 2014 at 21:16
Sidenote: This $pathOfFile = "/images/"; may fail. You may need to use a relative path. I.e.: $pathOfFile = "images/"; or $pathOfFile = "../images/"; — Funk Forty Niner
– Funk Forty Niner, Commented Jan 16, 2014 at 21:18

R3tep · Accepted Answer · 2014-01-16 21:27:24Z

2

You don't return a default value in your function. Maybe it's the problem :

function fileUploaded() {
    $fileName = $_FILES ['picture'] ['name'];
    $pathOfFile = "/images/";
    $fileTmpLoc = $_FILES ['picture'] ["tmp_name"];
    $fileResult = move_uploaded_file ( $fileTmpLoc, $pathOfFile );
    if (isset ( $fileName )) {
        return true;
    }
    return false;
}

answered Jan 16, 2014 at 21:27

R3tep

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1 Comment

ekrem777 Over a year ago

Again the same error. I will just delete the function and do it within main php with the help of if statement.

Gasim · Accepted Answer · 2014-01-16 21:35:06Z

//functions.php
function fileUpload($path) {
    if(!isset($_FILES['picture'])) return false;
    $fileName = $_FILES['picture']['name'];
    $fileTmpLoc = $_FILES['picture']['tmp_name'];
    if(move_uploaded_file ($fileTmpLoc, $path)) return $fileName;
    return false;
}

//main.php
include('functions.php');

$fileName = fileUpload('/images/');
if($fileName === false) {
   $fileName = 'default.jpg'; 
}

//do the rest here

Something similar to the above code. Since your function is in a different file, you need to include it (or require it)

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