因为这似乎是我必须做的才能获得这种效果:
$arr = ['a'=>'first', 'b'=>'second', ...];
$iter = new ArrayIterator( $arr );
// Do a bunch of iterations...
$iter->next();
// ...
$new_iter = new ArrayIterator( $arr );
while( $new_iter->key() != $iter->key() ) {
$new_iter->next();
}
编辑:此外,为了清楚起见,我不应该使用 unset()
修改基本数组吗?我认为数组迭代器存储它自己的基本数组副本,因此使用 offsetUnset()
似乎不正确。
最佳答案
ArrayIterator
没有实现 tell()
函数,但您可以模拟它,然后使用 seek()
去你想要的位置。这是一个扩展类,它就是这样做的:
<?php
class ArrayIteratorTellable extends ArrayIterator {
private $position = 0;
public function next() {
$this->position++;
parent::next();
}
public function rewind() {
$this->position = 0;
parent::rewind();
}
public function seek($position) {
$this->position = $position;
parent::seek($position);
}
public function tell() {
return $this->position;
}
public function copy() {
$clone = clone $this;
$clone->seek($this->tell());
return $clone;
}
}
?>
使用:
<?php
$arr = array('a' => 'first', 'b' => 'second', 'c' => 'third', 'd' => 'fourth');
$iter = new ArrayIteratorTellable( $arr );
$iter->next();
$new_iter = new ArrayIteratorTellable( $arr );
var_dump($iter->current()); //string(6) "second"
var_dump($new_iter->current()); //string(6) "first"
$new_iter->seek($iter->tell()); //Set the pointer to the same as $iter
var_dump($new_iter->current()); //string(6) "second"
?>
或者,您可以使用自定义 copy()
函数克隆对象:
<?php
$arr = array('a' => 'first', 'b' => 'second', 'c' => 'third', 'd' => 'fourth');
$iter = new ArrayIteratorTellable( $arr );
$iter->next();
$new_iter = $iter->copy();
var_dump($iter->current()); //string(6) "second"
var_dump($new_iter->current()); //string(6) "second"
?>
关于php - 如何复制 ArrayIterator 以保留其当前迭代位置?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20068602/