I am having problems with user written functions. The following statements don't seem to work inside the function when I pass a filename as a parameter (value or reference).

$userfile_name
$userfile_type

example

function upload($file) {

if ($file_name) {
...}

if ($file_type) {
...}

} //end upload

...
upload($userfile);
...

Is there something unique with functions that reside in the same .php file ? I am testing on a workstation using omni httpd, is it related to this web server ? I am even having trouble returning modified variables in the function back to the caller. Not using return statement. If I pass by reference, the passed parameter should take on any changes in the function.

Help, programmer getting frustrated with PHP !!!

Originally posted by rrsnider

function upload($file) {

if ($file_name) {
...}

if ($file_type) {
...}

} //end upload

...
upload($userfile);
...


Seems to be a variable scope problem, you need to declare "outside" variables as global whenever you wish to use them in functions, maybe the following example will illustrate it better.

<?php
// These are the variables declared outside the scope
$a = 2;
$b = 3;
function doadd() {
global $a;
$b = 6;

echo "Entered Function<br>";
// This will print 8 and NOT 5
echo $a+b;
}

echo "Starting execution<br>";
// This will print 5
echo $a+b;
// This will print 8
doadd();
?>


Hope that helps :D

Yes, but I'm trying to accomplish this: (I am questioning the passed parameter) does not seem to work.

function upload($file) {

if ($file_name) {
echo "$file_name";
...}

if ($file_type) {
echo "$file_type";
...}

} //end upload

...
upload($userfile1);
upload($userfile2);
upload($userfile3);

you are just passing $file, you need to pass $file_name $file_type too if you want your function to process them, so it would be something like

function upload($file, $file_name, $file_type) {
.
.
.
}

upload($userfile1, $userfile1_name, $userfile1_type);
upload($userfile2, $userfile2_name, $userfile2_type);
upload($userfile3, $userfile3_name, $userfile3_type);

Thats what I ended up doing. However, what prevents me from just passing $file as parameter and referencing $file_name in the function ?

Originally posted by rrsnider
Thats what I ended up doing. However, what prevents me from just passing $file as parameter and referencing $file_name in the function ?

Why pass a reference? It wont save you any memory (probably couple of bytes), but you can do that too I guess, just add & beside the arguments

Originally posted by rrsnider
Thats what I ended up doing. However, what prevents me from just passing $file as parameter and referencing $file_name in the function ?

Actually, $file and $file_name are completely separate variables. They both have the prefix "file" in their name but are separate variables.

HTH

Originally posted by rrsnider
Thats what I ended up doing. However, what prevents me from just passing $file as parameter and referencing $file_name in the function ?

You would need to define it as:

function upload($file_name)
{
if($file_name)
{
print $file_name;
}
}

upload($file);

This will work. But if you put $file up top then try an IF statement with $file_name, then like funkee said, two different variables.

You and also add & to the front of $file_name to make it a global variable so that:

function upload(&$file_name)

will make $file-name have the ability to be changed.