php json解码，如何获取该值？

下面是json数据，我想得到“average”值。我该怎么办

{

我使用函数json_decode来解析下面的json数据。 如何获得“平均”值

$jsonObj = json_decode($contentStr);
$items = $jsonObj->items;

---

必须重新格式化一些内容，不确定是无效的JSON还是外来字符，但这应该可以：

<?
          $js = '{ "items" : [{"city": "a",
             "tel": "85306308-1004",
             "name": "b",
             "mayor_id": 877861620,
             "venue_info": {"rating": 7.6,
                          "average": 116
                         },
             "lon": 116.437031906407,
             "checkin_users_num": 133,
             "lat": 39.9131007742324,
             "checkin_num": 161,
             "addr": "b",
             "dist": "200 m",
             "guid": "774E9ED4B79AF2A9904ECDA2F8D70565",
             "description": "c",
             "img": "http://img.jiepang.com/get/photo/182a6154e2dedbdf111bf29347ad6aa7?size=120"
            },
            {},
            {}],
    "num_items": 44}';

    $o = json_decode($js);

    $average = $o->items[0]->venue_info->average;

?>

打印（$items）

。这是显而易见的。

<?
          $js = '{ "items" : [{"city": "a",
             "tel": "85306308-1004",
             "name": "b",
             "mayor_id": 877861620,
             "venue_info": {"rating": 7.6,
                          "average": 116
                         },
             "lon": 116.437031906407,
             "checkin_users_num": 133,
             "lat": 39.9131007742324,
             "checkin_num": 161,
             "addr": "b",
             "dist": "200 m",
             "guid": "774E9ED4B79AF2A9904ECDA2F8D70565",
             "description": "c",
             "img": "http://img.jiepang.com/get/photo/182a6154e2dedbdf111bf29347ad6aa7?size=120"
            },
            {},
            {}],
    "num_items": 44}';

    $o = json_decode($js);

    $average = $o->items[0]->venue_info->average;

?>