PHP closure scope problem

问题

Apparently $pid is out of scope here. Shouldn't it be "closed" in with the function? I'm fairly sure that is how closures work in javascript for example.

According to some articles php closures are broken, so I cannot access this?

So how can $pid be accessed from this closure function?

class MyClass {
  static function getHdvdsCol($pid) {
    $col = new PointColumn();
    $col->key = $pid;
    $col->parser = function($row) {
        print $pid; // Undefined variable: pid
    };
    return $col;
  }
}

$func = MyClass::getHdvdsCol(45);
call_user_func($func, $row);

Edit I have gotten around it with use: $col->parser = function($row) use($pid). However I feel this is ugly.

回答1:

You need to specify which variables should be closed in this way:

function($row) use ($pid) { ... }

回答2:

You can use the bindTo method.

class MyClass {
  static function getHdvdsCol($pid) {
    $col = new PointColumn();
    $col->key = $pid;
    $parser = function($row) {
        print $this->key;
    };
    $col->parser = $parser->bindTo($parser, $parser);
    return $col;
  }
}

$func = MyClass::getHdvdsCol(45);
call_user_func($func, $row);

回答3:

I think PHP is very consistent in scoping of variables. The rule is, if a variable is defined outside a function, you must specify it explicitly. For lexical scope 'use' is used, for globals 'global' is used.

For example, you can't also use a global variable directly:

$n = 5;

function f()
{
    echo $n; // Undefined variable
}

You must use the global keyword:

$n = 5;

function f()
{
    global $n;
    echo $n;
}

来源：https://stackoverflow.com/questions/6543419/php-closure-scope-problem

标签

php

scope

closures