PHP closure scope problem

2020-02-25 22:39发布

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.

3条回答
神经病院院长
2楼-- · 2020-02-25 23:17

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);
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男人必须洒脱
3楼-- · 2020-02-25 23:27

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

function($row) use ($pid) { ... }
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Rolldiameter
4楼-- · 2020-02-25 23:35

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;
}
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