我们有两个PHP5对象,希望将其中一个对象的内容合并到另一个对象中。它们之间没有子类的概念,因此以下主题中描述的解决方案无法应用。
How do you copy a PHP object into a different object type
//We have this:
$objectA->a;
$objectA->b;
$objectB->c;
$objectB->d;
//We want the easiest way to get:
$objectC->a;
$objectC->b;
$objectC->c;
$objectC->d;
备注:
发布于 2009-01-18 19:18:16
foreach($objectA as $k => $v) $objectB->$k = $v;
发布于 2009-01-18 19:17:34
您可以创建另一个对象,将对魔术方法的调用分派到底层对象。下面是处理__get
的方法,但要让它完全工作,您必须覆盖所有相关的魔术方法。你可能会发现语法错误,因为我只是在脑海中输入了它。
class Compositor {
private $obj_a;
private $obj_b;
public function __construct($obj_a, $obj_b) {
$this->obj_a = $obj_a;
$this->obj_b = $obj_b;
}
public function __get($attrib_name) {
if ($this->obj_a->$attrib_name) {
return $this->obj_a->$attrib_name;
} else {
return $this->obj_b->$attrib_name;
}
}
}
祝好运。
发布于 2009-05-27 22:12:28
我知道使用泛型对象stdClass()并将它们转换为数组可以回答这个问题,但我认为合成器是一个很好的答案。然而,我觉得它可以使用一些功能增强,并可能对其他人有用。
功能:
obj组合指定引用或克隆合并指定第一个或最后一个条目采用precedence
(...)
代码:
class Compositor {
protected $composite = array();
protected $use_reference;
protected $first_precedence;
/**
* __construct, Constructor
*
* Used to set options.
*
* @param bool $use_reference whether to use a reference (TRUE) or to copy the object (FALSE) [default]
* @param bool $first_precedence whether the first entry takes precedence (TRUE) or last entry takes precedence (FALSE) [default]
*/
public function __construct($use_reference = FALSE, $first_precedence = FALSE) {
// Use a reference
$this->use_reference = $use_reference === TRUE ? TRUE : FALSE;
$this->first_precedence = $first_precedence === TRUE ? TRUE : FALSE;
}
/**
* Merge, used to merge multiple objects stored in an array
*
* This is used to *start* the merge or to merge an array of objects.
* It is not needed to start the merge, but visually is nice.
*
* @param object[]|object $objects array of objects to merge or a single object
* @return object the instance to enable linking
*/
public function & merge() {
$objects = func_get_args();
// Each object
foreach($objects as &$object) $this->with($object);
// Garbage collection
unset($object);
// Return $this instance
return $this;
}
/**
* With, used to merge a singluar object
*
* Used to add an object to the composition
*
* @param object $object an object to merge
* @return object the instance to enable linking
*/
public function & with(&$object) {
// An object
if(is_object($object)) {
// Reference
if($this->use_reference) {
if($this->first_precedence) array_push($this->composite, $object);
else array_unshift($this->composite, $object);
}
// Clone
else {
if($this->first_precedence) array_push($this->composite, clone $object);
else array_unshift($this->composite, clone $object);
}
}
// Return $this instance
return $this;
}
/**
* __get, retrieves the psudo merged object
*
* @param string $name name of the variable in the object
* @return mixed returns a reference to the requested variable
*
*/
public function & __get($name) {
$return = NULL;
foreach($this->composite as &$object) {
if(isset($object->$name)) {
$return =& $object->$name;
break;
}
}
// Garbage collection
unset($object);
return $return;
}
}
用法:
$obj = new Compositor(use_reference, first_precedence);
$obj->merge([object $object [, object $object [, object $...]]]);
$obj->with([object $object]);
示例:
$obj1 = new stdClass();
$obj1->a = 'obj1:a';
$obj1->b = 'obj1:b';
$obj1->c = 'obj1:c';
$obj2 = new stdClass();
$obj2->a = 'obj2:a';
$obj2->b = 'obj2:b';
$obj2->d = 'obj2:d';
$obj3 = new Compositor();
$obj3->merge($obj1, $obj2);
$obj1->c = '#obj1:c';
var_dump($obj3->a, $obj3->b, $obj3->c, $obj3->d);
// obj2:a, obj2:b, obj1:c, obj2:d
$obj1->c;
$obj3 = new Compositor(TRUE);
$obj3->merge($obj1)->with($obj2);
$obj1->c = '#obj1:c';
var_dump($obj3->a, $obj3->b, $obj3->c, $obj3->d);
// obj1:a, obj1:b, obj1:c, obj2:d
$obj1->c = 'obj1:c';
$obj3 = new Compositor(FALSE, TRUE);
$obj3->with($obj1)->with($obj2);
$obj1->c = '#obj1:c';
var_dump($obj3->a, $obj3->b, $obj3->c, $obj3->d);
// obj1:a, obj1:b, #obj1:c, obj2:d
$obj1->c = 'obj1:c';
https://stackoverflow.com/questions/455700
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