I have file a.php that looks like
function func_1() {
inlude_once(b.php);
$somevar = 'def';
func_2($somevar);
}
and b.php that looks like
$some_global_var = 'abc';
function func_2($var) {
global $some_global_var;
echo $some_global_var.$var;
}
And for some reason I get only def as a result, why func_2 don't see $some_global_var ?
Because you forgot the scope of func_1. So when you include your define this is how your code appears to PHP
function func_1() {
$some_global_var = 'abc'; // <- this is inside the scope of the parent function!
function func_2($var) {
global $some_global_var;
echo $some_global_var.$var;
}
$somevar = 'def';
func_2($somevar);
}
You're doing it inside func_1. So the variable was never really available in the global scope. If you defined $some_global_var = 'abc'; outside, then it's in the global scope.
What you should do is inject this as an argument instead. Globals are a bad practice
function func_1() {
$some_global_var = 'abc';
function func_2($var, $var2) {
echo $var2 . $var;
}
$somevar = 'def';
func_2($somevar, $some_global_var);
}
put global in front of it.
per the PHP docs
Using global keyword outside a function is not an error. It can be used if the file is included from inside a function.
you may have an issue with an include file getting in the way
global the variable in the include it doesn't solve the problemParse error: syntax error, unexpected '=', expecting ',' or ';' in if I global the variable in the include