Не решить в кодировании JSON [закрыто] ⇐ Php

[Anonymous]

Я не хочу собирать все данные — только данные об устройстве, группе и датчике. Затем я отправлю его на другой сервер.
Пожалуйста, помогите мне решить это на php:
data:
{"items":
[
{"id":1,"uniqueId":"1234","name":"Lamp","group": {"id":1,"title":"A000"},"stop_duration_sec":35629,"sensors":[ {"id":4,"type":"Panel","name":"Room"}, {"id":33,"type":"Switch","name":"On"}]},
{"id":2,"uniqueId":"456","name":"Fan","group":{"id":5,"title":"A1000"},"stop_duration_sec":35629,"sensors":[{"id":4,"type":"Panel","name":"Room"}, {"id":33,"type":"Switch","name":"On"}]}
]
}

Мой код на php:
$data = json_decode($json, TRUE);

foreach($data as $key) {
`$group=$key['id'];`
$title=$key['title'];

foreach($key['items'] as $items) {
$id=$items['id'];
$uniqueId=$items['uniqueId'];
$devicename=$items['name'];
$stop_duration_sec=$items['stop_duration_sec'];
foreach($items['sensors'] as $sensors) {
$sensorid=$sensors['id'];
$type=$sensors['type'];
$sensorname=$sensors['name'];
}

$data['items'] = array('id' =>$id,
'uniqueId' => $uniqueId,
'name' => $devicename,
'group' => (array(
'id' => $group,
'title'=> $title,)),
'stop_duration_sec' => $stop_duration_sec,
'sensor' => (array(
'id' => $sensorid,
'type' => $sensors,
'name'=> $sensorname,)),
);

}
echo json_encode($data);

}
?>


Подробнее здесь: https://stackoverflow.com/questions/793 ... ncode-json