variable define error in PHP picture uploading [duplicate] - php

Question

This question already has answers here:
What does the PHP error message "Notice: Use of undefined constant" mean?
(2 answers)
Closed 8 years ago.
I'm only a beginner with PHP and server side programming so excuse me for this question about variables. I was reading a tutorial about uploading a photo.
(tutorial on plus2net)
and like described in the tutorial i directly copied and pasted the code after understanding it. However the webpage filled with following errors after i inserted PHP in to the HTML.
Notice: Use of undefined constant size - assumed 'size' in C:\Program Files (x86)\XAMPP\htdocs\college_ink\shirtDesign.php on line 44
Notice: Undefined index: file_up in C:\Program Files
(x86)\XAMPP\htdocs\college_ink\shirtDesign.php on line 44
Notice: Use of undefined constant file_up - assumed 'file_up' in
C:\Program Files (x86)\XAMPP\htdocs\college_ink\shirtDesign.php on
line 45
Notice: Use of undefined constant name - assumed 'name' in C:\Program
Files (x86)\XAMPP\htdocs\college_ink\shirtDesign.php on line 45
Notice: Undefined index: file_up in C:\Program Files
(x86)\XAMPP\htdocs\college_ink\shirtDesign.php on line 45
Notice: Use of undefined constant file_up - assumed 'file_up' in
C:\Program Files (x86)\XAMPP\htdocs\college_ink\shirtDesign.php on
line 47
Notice: Use of undefined constant size - assumed 'size' in C:\Program
Files (x86)\XAMPP\htdocs\college_ink\shirtDesign.php on line 47
Notice: Undefined index: file_up in C:\Program Files
(x86)\XAMPP\htdocs\college_ink\shirtDesign.php on line 47
Notice: Use of undefined constant file_up - assumed 'file_up' in
C:\Program Files (x86)\XAMPP\htdocs\college_ink\shirtDesign.php on
line 51
This is the PHP Code:
<?Php
$file_upload="true";
$file_up_size=$_FILES['file_up'][size];
echo $_FILES[file_up][name];
if ($_FILES[file_up][size]>500000){
$msg=$msg."Uploaded file size more than 500KB<BR>";
$file_upload="false";}
if (!($_FILES[file_up][type] =="image/jpeg" OR $_FILES[file_up][type] =="image/png")){
$msg=$msg."Your uploaded file must be of JPG or PNG.<BR>";
$file_upload="false";}
$file_name=$_FILES[file_up][name];
$add="upload/$file_name"; //the path with the file name
if($file_upload=="true"){
if(move_uploaded_file ($_FILES[file_up][tmp_name], $add)){
//do your coding here to give a thanks message
}
else{
echo "Failed to upload file.";}
}
else{
echo $msg;
}
?>
I've corrected the variable error but I'm getting this error now instead
Notice: Undefined index: file_up in C:\Program Files
(x86)\XAMPP\htdocs\college_ink\shirtDesign.php on line 44
Please guide me through this. :)

Answer 1

Where you have
$file_up_size=$_FILES['file_up'][size];
size is being interpreted as a constant due to the lack of a dollar sign.
What you want is
$file_up_size=$_FILES['file_up']['size'];

Answer 2

The problem is that you are relying on bare strings being interpreted as such. Normally, in PHP code, this is considered a constant:
echo FOO;
If you don't define that constant first, like this:
define('FOO', 'foo');
You will encounter this error and PHP will interpret it as a bare string and echo FOO. This is a bad choice. I think what you mean to do is use actual strings (rather than undefined constants) as your array indices:
if ($_FILES['file_up']['size']>500000){
// etc..
Which is what is recommended.
Note: Bare strings are covered in the manual entry for Arrays, specifically underneath the section Arrays do's and don'ts (the manual says don't do this).
This is wrong, but it works. The reason is that this code has an
undefined constant (bar) rather than a string ('bar' - notice the
quotes). PHP may in future define constants which, unfortunately for
such code, have the same name. It works because PHP automatically
converts a bare string (an unquoted string which does not correspond
to any known symbol) into a string which contains the bare string. For
instance, if there is no defined constant named bar, then PHP will
substitute in the string 'bar' and use that.