在PHP中的函数之间传递变量

Question

I'm slowly converting my work mode to functions and as well, OOP.

I have a file, functions.php which I call at the top of a log in page.

When the form on the page is submitted, I use a function within the functions file to check details entered are correct. If not I do eg

if ($how_many<1){

    $error .= "<p>You typed the wrong email address and/or password.</p>";
    return $error;

}

I then have another function named anyerrors in the same functions file. This is:

function anyerrors($error){
    echo $error;
}

UPDATE: Then in the file I call anyerrors($error); where I want the error to be shown.

But nothing is being shown? I'm guessing it's because I'm creating the variable in one function and it's not getting to the other function.

Am I way off in my logic?

For the record I'm an old school coder I guess and taking time to change my work flow radically one step at a time.

Answer 1

Your $error variable scope is limited to place where you declare.it can be in the file at global level or inside a function. function variables are primitive data types the life span is only till function exists. So if you declare $error in a function it will be a new variable. If you want to use a global variable you will have to call it as global.

I would personally avoid using global but it will work..

read about globals

$error = "";

if ($how_many<1){

    $error .= "<p>You typed the wrong email address and/or password.</p>";
    return $error;

}

function anyerrors($error){
    global $error;
    echo $error;
}

I would suggest create class which will make things easier.

Answer 2

it's a wrong way to use it switch the echo in the function with return and the echo to the function like this :

function anyerrors($error){
    return $error;
}



if ($how_many<1){

    $error .= "<p>You typed the wrong email address and/or password.</p>";
    echo anyerrors($error);

}

or to keep it simple you can just print it after creating the error

if ($how_many<1){

    $error .= "<p>You typed the wrong email address and/or password.</p>";
    echo $error;

}

the answer above this just to give you right way to use function that return a value I hope my answer can help you and if you have another question about my answer feel free to ask in the comment :)

Answer 3

Looks like lackluster code it should be something like this

function checkForErrors($type)
    if ($type =='your type you want')
    {
        if ($how_many<1){

            $error .= "<p>You typed the wrong email address and/or password.</p>";
            return $error;
        }

    }
}
function anyerrors(){
    $errors = checkForErrors('your type you want');
    if($errors!="") {
          echo $errors;
    }
}