php - How to pass parameters to a function

Question

I'm pretty sure this is a very basic question to all of you, but i'm new with php, and i don't really get it... basically i've created a function in which i need to pass two parameters.

My functions is this:

function displayRoomDetails($customerRooms, $test)
{
    foreach ($customerRooms as $room) {
        $test.= $room->name;
    };
}

it is a very basic function, but will do for this example.

Now, i'm creating templates to display several information, and i have 3 different layout where i need to display the same info but styled differently, so my approach was:

template1 .= '<span>';

if (!$customerRooms == "") {
    displayRoomDetails($customerRooms,"template1");
};

template1 .= '</span>';

It should be pretty easy to understand, basically i'm calling the same functions in all the different templates passing as a parameter the template name and trying to append the results to the right template.

The problem i've got is this: According to this example here -> http://www.w3schools.com/php/showphp.asp?filename=demo_function3

i should be able to do this exactly like i did, but when i try, when i debug my function, $template doesn't take the passed value as i though it would, but it is still called $test and not $template1...

What am i doing wrong?

Thanks

The example you link to has practically nothing in common with your code at all. — Quentin
– Quentin, Commented Oct 21, 2015 at 10:17
Your code also immediately throws an error: PHP Parse error: syntax error, unexpected '.=' (T_CONCAT_EQUAL) in - on line 2 — Quentin
– Quentin, Commented Oct 21, 2015 at 10:18
its like you are calling "template1" within template1 and you want to display "template1".. if am not worng. — Sagar Pawar
– Sagar Pawar, Commented Oct 21, 2015 at 10:21
that if statement isn't doing what you think it is if (!$customerRooms == "") the variable is being cast to boolean and inverted but the result is not being assigned to anything. You're still comparing $customerRooms == "" you probably need if($customerRooms !== "") — andrew
– andrew, Commented Oct 21, 2015 at 10:23

Lars Hansen · Accepted Answer · 2015-10-21 10:26:32Z

2

Try these changes:

function displayRoomDetails($customerRooms, &$test)

And

$template1 .= '<span>';
if ($customerRooms != "") {
  displayRoomDetails($customerRooms, $template1);
};
$template1 .= '</span>';

answered Oct 21, 2015 at 10:26

Lars Hansen

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4 Comments

andrew Over a year ago

You should explain why you suggested using the &operator and what it does

Nick Over a year ago

Thanks! It's working. This is exactly what i needed.

alanmm Over a year ago

Another copy-and-paste solution.

Lars Hansen Over a year ago

@andrew - yep, you are right. Nick - do you know what the & means?

Hipolith · Accepted Answer · 2015-10-21 10:32:43Z

From what I understand you want to append some text to the template1 variable using the displayRoomDetailsFunction

Some things to fix:

template1 should be $template1
You should be passing the $template1 not the "template1" (i.e. the variable itself not its name).
If you want to modify this variable you need to either:
- pass it as reference, which you can do by changing the function's declaration to: function displayRoomDetails($customerRooms, &$test)
- return new string from function and assign it to the $template1 by adding return $test; just after your foreach block and changing the call to $template1 .= displayRoomDetails($customerRooms,$template1);

Additional note: if $customerRooms is an array, it'd be better to check if it's not empty using count() than !$customerRooms == "", see @andrew's comment for details

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