PHP require_once在我的代码中不起作用

Question

this is probably a basic question about php, but I can't figure out how to solve it.

Well, I have a file called globals.php

<?php
$DATA = array(
    'first' => 'LOL',
    'second' => 'Whatever'
);
?>

And I also have another file (omg.php) with the following function:

<?php

require_once('globals.php');

function print_text_omg($selector = 0){
global $DATA; //added this line of code. NOW IT WORKS
$var = '';
if($selector == 0){
    $var = '';
}else{
    $var = 'Hi. ';
}

//$DATA is a variable from globals.php that is supposed to be declared in require_once('globals.php');
//$var is a variable inside the function print_text_omg
//I am trying to concatenate string $var with the string $DATA['first']

$finaltext = $var.$DATA['first'];
echo $finaltext;
}
?>

Then, in main.php I have this:

<?php 
include('omg.php');
print_text_omg();
print_text_omg(1);
?>

This should print something like:

//LOL
//Hi. LOL

Instead, I have this warning:

Notice: Undefined variable: DATA in ...

Which is the part of $finaltext = $var.$DATA['first'];

UPDATE

Thanks to user Casimir et Hippolyte' suggestion, I've edited my function and it works now. Added the line that worked for me.

Answer 1

It doesn't work because $DATA isn't in the scope of your function.

To make $DATA available inside your function, you must pass it as a parameter to the function or define $DATA as a global variable.

The problem isn't related to require_once .

http://php.net/manual/en/language.variables.scope.php

Answer 2

The error is correct, first I can't find where did you declare $var (maybe it should be an object of some class?) and also $var doesn't contain the $DATA array, and for last, . operator in php is for concatenate strings, for object navigation is -> operator

Answer 3

You haven't declared the variable $var , plus . is php's concatenation operator. Try changing your code to this

$finaltext = $DATA['first'] ;