php lessons/practice. need help understanding if/else - php

Question

ok so im here trying to practice some php (im a super beginner) and so long story short,
i put form elements in one page, passed it to the process php.
Im just messing aound trying to see what ive learned so far. i dont get any errors, just dont understand why it doesnt work.
<?php
$yourname = htmlspecialchars($_POST['name']);
$compname = htmlspecialchars($_POST['compName']);
$response = array("please enter correct information","hmm" . "$yourname");
function nametest() {
if (!isset($yourname)){
$yourname = $response[0];}
else {
$yourname = $response[1];;
}
}
?>
<?php nametest(); ?>
what im trying to do is, that if the name isnt set, to make a variable equal to a value inside response.

Answer 1

Try
function nametest() {
if (!isset($yourname)){
$yourname = $response[0];
} else {
$yourname = $response[1];
}
return $yourname;
}
print nametest();
The function needs to return a value to be printed. I also noticed you have two ;; behind line 5.

Answer 2

Because you are assigning $yourname and $compname in the first two lines:
$yourname = htmlspecialchars($_POST['name']);
$compname = htmlspecialchars($_POST['compName']);
UPDATE You can check if these are set in POST, and therefore not need to check them later:
$yourname = isset($_POST['name']) ? htmlspecialchars($_POST['name']) : "oops, no value";
$compname = isset($_POST['compName']) ? htmlspecialchars($_POST['compName']) : "oops, no value";
They will always be set, even if NULL or empty. So, your later calls to isset() will always be true.
Instead, you may check if they are empty with the empty() function:
UPDATE Not necessary according to corrections in comments. Your isset() should work.
// Check with empty()
// but still won't work properly. keep reading below...
function nametest() {
if (!empty($yourname)){
$yourname = $response[0];}
else {
$yourname = $response[1];;
}
}
However, there is another problem here of variable scope. The variables are not available inside the function unless you either pass them in as parameters or use the global keyword:
// $yourname is passed as a function argument.
function nametest($yourname, $response) {
if (!empty($yourname)){
$yourname = $response[0];}
else {
$yourname = $response[1];;
}
}
Getting there... Now your function assigns $yourname, but it doesn't return or print any value. Add a return statement, and then you can echo out the result:
function nametest($yourname, $response) {
if (!empty($yourname)){
$yourname = $response[0];}
else {
$yourname = $response[1];;
}
// Add a return statement
return $yourname;
}
// Now call the function, echo'ing its return value
echo nametest($yourname, $response);

Answer 3

Variable Scope is the biggest mistake here, your function can not 'see' the variables that you created outside of it, do this:
<?php
.
.
.
function nametest($yourname, $response) { // This creates two new variables that
// are visible only by this function
if (!isset($yourname)){
$yourname = $response[0];
} else {
$yourname = $response[1]; // Get rid of the extra semicolon
}
return $yourname; // This $yourname is only visible by this function so you
// need to send it's value back to the calling code
}
?>
<?php nametest($yourname, $response); ?> // This sends the values of the
// variables that you created at the
// top of this script