xml转json,怎么筛选数据？

JSON XML 下边的xml，想转为json
但P节点下只需要这两个节点的数据，其他数据不需要。
如果直接用json_encode转出来，是所有的数据
请问怎么能转为json,但只要的数据，怎么弄？

例如
{"PN":"\u7b2c\u4e00\u5468\u64ad\u5267\u573a\uff1a\u8ffd\u9c7c\u4f20\u5947 31","PT":"2013-08-19 22:01:00"}

-------xml---------

110171675
张三
2013-08-19 00:02:00
46

24
74750
0
2013-08-19 00:02:00
AM

110171676
我是大美人
2013-08-19 01:15:00
46

24
74501
0
2013-08-19 01:15:00
AM

110171677
李四
2013-08-19 02:09:00
46

24
64519
71411
2013-08-19 02:09:00
AM

回复讨论(解决方案)

$string =   
    110171675张三2013-08-19 00:02:0046
24    7475002013-08-19 00:02:00AM
    110171676我是大美人2013-08-19 01:15:0046
24    7450102013-08-19 01:15:00AM
    110171677李四2013-08-19 02:09:0046
24    64519714112013-08-19 02:09:00AMXML;$xml = simplexml_load_string($string);foreach($xml->P as $item){     $item=(array)$item;    $arr[]=array('PN'=>$item['PN'],'PT'=>$item['PT']);}echo json_encode($arr);

[{"PN":"\u5f20\u4e09","PT":"2013-08-19 00:02:00"},{"PN":"\u6211\u662f\u5927\u7f8e\u4eba","PT":"2013-08-19 01:15:00"},{"PN":"\u674e\u56db","PT":"2013-08-19 02:09:00"}]

$xml =  
    110171675张三2013-08-19 00:02:0046
24    7475002013-08-19 00:02:00AM
    110171676我是大美人2013-08-19 01:15:0046
24    7450102013-08-19 01:15:00AM
    110171677李四2013-08-19 02:09:0046
24    64519714112013-08-19 02:09:00AMXML;$sm = simplexml_load_string($xml);foreach($sm->P as $item) {  $r[] = array('PN' => strval($item->PN), 'PT' => strval($item->PT));}echo json_encode($r);

[{"PN":"\u5f20\u4e09","PT":"2013-08-19 00:02:00"},{"PN":"\u6211\u662f\u5927\u7f8e\u4eba","PT":"2013-08-19 01:15:00"},{"PN":"\u674e\u56db","PT":"2013-08-19 02:09:00"}]