Passing variables between functions in PHP

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 e.g.

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.

Show how you're calling the function that returns $error. You need to assign the result to something. "old school coder"? Calling functions and returning values from them is the way people have been programming for decades. — Barmar
– Barmar, Commented May 12, 2013 at 9:40
Its in the post headed update: Then in the file I call anyerrors($error); where I want the error to be shown. — StudioTime
– StudioTime, Commented May 12, 2013 at 9:42
Is if($how_many<1) inside a function? It must be, since it uses return. Does the caller of that function assign its value to $error? — Barmar
– Barmar, Commented May 12, 2013 at 9:43

Dinesh · Accepted Answer · 2013-05-12 09:47:29Z

3

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.

edited May 12, 2013 at 9:47

answered May 12, 2013 at 9:42

Dinesh

3,1031 gold badge21 silver badges19 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

StudioTime Over a year ago

Thanks, works great. I'm reading more and more about OOP stuff and this is a great help to push me more towards it, thanks again.

Niko Adrianus Yuwono · Accepted Answer · 2013-05-12 09:27:39Z

0

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 :)

answered May 12, 2013 at 9:27

Niko Adrianus Yuwono

11.1k8 gold badges44 silver badges64 bronze badges

2 Comments

StudioTime Over a year ago

Thanks. I know I can use second method but that doesn't let me place the error where I want it to be shown unless I add that logic to the html file which I'm trying to avoid to keep the code clean and separated.

Niko Adrianus Yuwono Over a year ago

ah if the answer how to seperate your php and html file maybe you should take a look in this answer -> stackoverflow.com/a/95027/1545363

chandresh_cool · Accepted Answer · 2013-05-12 09:43:50Z

0

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;
    }
}

edited May 12, 2013 at 9:43

answered May 12, 2013 at 9:28

chandresh_cool

11.9k3 gold badges32 silver badges46 bronze badges

2 Comments

StudioTime Over a year ago

That would mean I create a new function for every single error check throughout the site, whereas what I want to do is simply populate variable $error within functions and then have one function to echo it if it exists. Is that not possible?

chandresh_cool Over a year ago

No you can pass argument in first function and make it generic and check for that particular error. Check my updated answer

Collectives™ on Stack Overflow

Passing variables between functions in PHP

3 Answers 3

1 Comment

2 Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

2 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related