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我目前有问题产生,因为我不使用PHP的那么多,但任何人都可以帮助我建立JSON象下面这样的JSON响应:如何在php中添加对象内的对象?
{
"set_attributes":
{
"apikey": "some value",
},
"block_names": ["getdata"],
"type": "show_block",
"title": "go"
}
这里是我的代码:
$response = array();
$response['block_names'] = array();
$response['block_names'] = "getdata";
$response['type'] = "show_block";
$response['title'] = "go";
if ($db->studentLogin($username, $password)) {
$student = $db->getStudent($username);
$temp = array();
$temp['apikey'] = $student['api_key'];
$response['set_attributes'] = $temp['apikey'];
} else {
$temp = array();
$response['messages'] = array();
$temp['text'] = "Invalid username or password";
array_push($response['messages'],$temp);
}
echoResponse(200, $response);
这表明像这样:
{
"block_names": "getdata",
"type": "show_block",
"title": "go",
"set_attributes": "f74911b29778adea86aa24d5ce85ff58"
}
但我想补充apikey = “f74911b29778adea86aa24d5ce85ff58” 内set_attributes和[]外block_names就像obove。我怎样才能做到这一点?
'$回应[ 'set_attributes'] = $温度;' - 现在你需要正义价值 – splash58
'$回应[ 'block_names'] =阵列(”的GetData“);' – splash58