我有一个返回下面的JSON对象的网址:从外部URL获取JSON对象与PHP
{
"addressList": {
"addresses": [
{
"id": 0000000,
"receiverName": "Name Example",
"country": {
"id": "BRA",
"name": "Brasil"
},
"state": {
"id": "SP"
},
"city": "São Paulo",
"zipcode": "00000000",
"type": "Residential",
"street": "000, St Example",
"number": 00,
"neighborhood": "Example",
"hash": "1bf09357",
"defaultAddress": false,
"notReceiver": false
}
]
}
}
我想要得到的state
值,我怎么能检索与PHP?
我试过了,就像这样,但我无法获得state
的值,在这种情况下应该是SP
。
$string = '{ "addressList": { "addresses": [ { "id": xxxxxx, "receiverName": "XXXXX XXXXX", "country": { "id": "BRA", "name": "Brasil" }, "state": { "id": "SP" }, "city": "São Paulo", "zipcode": "03164xxx", "type": "Residential", "street": "Rua xxx", "number": xx, "neighborhood": "xxxxx", "hash": "xxxxx", "defaultAddress": false, "notReceiver": false } ] } }';
$json_o = json_decode($string);
$estado = $json_o->state;
我怎样才能达到我想要的结果?
其AddressList中,地址(第一指标),状态,那么ID,你不能只是直接使用状态,其内部没有下父级别,请使用'print_r/var_dump'查找 – Ghost
请注意,JSON中的数字“八进制和十六进制格式未使用”。 – MinhTri