Php function calling, Undefined function error - php

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.

Answer 1

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;
}

Answer 2

//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)