PHP – use outside function with if-statements inside foreach-loop [duplicate] - php

Question

This question already has answers here:
Reference: What is variable scope, which variables are accessible from where and what are "undefined variable" errors?
(3 answers)
Closed 4 years ago.
I'm new to PHP, please be gentle.
What do I need to change to make this work in PHP?
<div>some HTML here</div>
<?php
function typ__1() {
if ($temperature >= 29) {
$hot = true;
} else {
$hot = false;
}
}
?>
<?php foreach (array_slice($data->something->something, 0, 5) as $day):
$temperature = $day->temperature;
typ__1();
if ($hot == true) {
$bottom = "Shorts";
} else if ($hot == false) {
$bottom = "Pants";
}
<div><?php echo $bottom ?></div>
<?php endforeach ?>
So the main issue/question is if I'm using the function correctly. Can I write if-statements in an outside function and then use them inside a
foreach-loop? The reason/goal is to shorten the foreach-loop.
(This is a reduced example, so there could be a typo somewhere in there.)
Thanks for your help!

Answer 1

Everything is about scope of PHP variables. You should "inject" variable into function like this:
<div>some HTML here</div>
<?php
function typ__1($temperature) {
if ($temperature >= 29) {
return true;
}
return false;
}
?>
<?php foreach (array_slice($data->something->something, 0, 5) as $day):
if (typ__1($day->temperature)) {
$bottom = "Shorts";
} else if (typ__1($day->temperature)) {
$bottom = "Pants";
}
<div><?php echo $bottom ?></div>
<?php endforeach ?>
http://php.net/manual/en/language.variables.scope.php

Answer 2

Add parameter in your function and return a value.
<?php
function typ__1($temperature) {
if ($temperature >= 29) {
$hot = true;
} else {
$hot = false;
}
return $hot;
}
?>
<?php foreach (array_slice($data->something->something, 0, 5) as $day):
$temperature = $day->temperature;
$hot=typ__1($temperature);
if ($hot == true) {
$bottom = "Shorts";
} else if ($hot == false) {
$bottom = "Pants";
}
<div><?php echo $bottom ?></div>
<?php endforeach ?>