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我正在检索一个JSON响应,除了一个部分外没有任何问题。有些项目是嵌套一个JSONObjects:如何从嵌套的JSON响应中检索信息?
{
"response": {
"venues": [
{
"id": "42829c80f964a5205f221fe3",
"name": "AmericanExpressTower",
"contact": {
"phone": "2126405130",
"formattedPhone": "(212)640-5130",
"twitter": "americanexpress"
},
"location": {
"address": "200VeseySt",
"crossStreet": "WestSt",
"lat": 40.713618978735,
"lng": -74.01408649911748,
"distance": 1926,
"postalCode": "10285",
"city": "NewYork",
"state": "NY",
"country": "UnitedStates"
}
}
]
}
}
我怎样才能像 “formattedPhone” 接触对象访问一个项目?
我可以访问“名称”每个项目的罚款:
JSONObject json = new JSONObject(result);
JSONArray venues = json.getJSONObject("response")
.getJSONArray("venues");
input.close();
int vLength = venues.length();
StringBuilder builder = new StringBuilder();
for (int i = 0; i < vLength; i++) {
builder.append("Location: ");
builder.append(venues.getJSONObject(i)
.getString("name").toString());
builder.append("\n");
}
我还要补充一点,这样做:JSONObject contacts = new JSONObject(venues.getJSONObject(i).getString(“contact”));将拉动整个对象。 – Paul
因此contacts.getString(“格式化的电话”)从上面? –
我试过了,它说没有格式化电话 – Paul