【发布时间】:2018-10-28 12:23:08
【问题描述】:
我有 REST API,如下所示:
{
"rajaongkir": {
"query": {
"key": "b5231ee43b8ee75764bd6a289c4c5745"
},
"status": {
"code": 200,
"description": "OK"
},
"results": [
{
"city_id": "1",
"province_id": "21",
"province": "Nanggroe Aceh Darussalam (NAD)",
"type": "Kabupaten",
"city_name": "Aceh Barat",
"postal_code": "23681"
},
{
"city_id": "2",
"province_id": "21",
"province": "Nanggroe Aceh Darussalam (NAD)",
"type": "Kabupaten",
"city_name": "Aceh Barat Daya",
"postal_code": "23764"
}
]
}
}
我想使用这个 API。我将对象称为以下代码:
<!doctype html>
<html lang="en">
<head>
<!-- Required meta tags -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1, shrink-to-fit=no">
<!-- Bootstrap CSS -->
<link rel="stylesheet" href="https://stackpath.bootstrapcdn.com/bootstrap/4.1.3/css/bootstrap.min.css" integrity="sha384-MCw98/SFnGE8fJT3GXwEOngsV7Zt27NXFoaoApmYm81iuXoPkFOJwJ8ERdknLPMO" crossorigin="anonymous">
<title>Raja Ongkir</title>
</head>
<body>
<h1>Raja Ongkir</h1>
<?php
echo var_dump($data->rajaongkir->results[0]);
?>
</body>
</html>
如果我使用这一行调用 JSON 的第一个元素
echo var_dump($data->rajaongkir->results[0]);
或
echo var_dump($data->rajaongkir->results[0]->city_name);
我得到了我想要的输出。但是,如果我尝试在 results 对象中获取所有 city_id 或 city_name,请使用此代码
echo var_dump($data->rajaongkir->results->city_name);
我收到了这个错误
消息:试图获取非对象的属性“city_name”
如何解决?
【问题讨论】:
标签: php json codeigniter