PHP变量生命周期/范围

Question

I'm a Java developer and recently I'v been tasked with PHP code review. While going through the PHP source code, I'v noticed that a variable is initialised in an if, while, switch and do statement then same variable is used outside these statements. Below is a fragment the code

Senario 1

if ($status == 200) {
     $messageCode = "SC001";
}

// Here, use the $message variable that is declared in an if
$queryDb->selectStatusCode($message);

Senario 2

foreach ($value->children() as $k => $v) {
    if ($k == "status") {
        $messageCode = $v;
    }
}

// Here, use the $messageCode variable that is declared in an foreach
$messageCode ....

In my understanding, a variable declared in a control statement is only accessible with in the control code block.

My question is, what is the variable scope of a variable in a PHP function and how is this variable accessible outside a control statement block?

How is that above code working and producing expected results?

Answer 1

In PHP control statements have no separate scope. They share the scope with the outer function or the global scope if no function is present. ( PHP: Variable scope ).

$foo = 'bar';

function foobar() {
    $foo = 'baz';

    // will output 'baz'
    echo $foo;
}

// will output 'bar'
echo $foo;

Your variables will have the last value assigned within the control structure. It is good practice to initialize the variable before the control structure but it is not required.

// it is good practice to declare the variable before
// to avoid undefined variables. but it is not required.
$foo = 'bar';
if (true == false) {
    $foo = 'baz';
}

// do something with $foo here

Namespaces do not affect variable scope. They only affect classes, interfaces, functions and constants ( PHP: Namespaces Overview ). The following code will output 'baz':

namespace A { 
    $foo = 'bar';
}

namespace B {
    // namespace does not affect variables
    // so previous value is overwritten
    $foo = 'baz';
}

namespace {
    // prints 'baz'
    echo $foo;
}