[英]Parsing JSON throws exception
我正在嘗試解析JSON以從中獲取一些字段。 以下是我的JSON-
{
"id": "100000224942080000",
"name": "Tech Geeky",
"first_name": "Tech",
"last_name": "Geeky",
"link": "https://www.facebook.com/tech.geeky",
"username": "tech.geeky",
"work": [
{
"employer": {
"id": "1854993931353456",
"name": "Tech"
},
"location": {
"id": "1119482345542155151",
"name": "Santa Cruz, California"
},
"position": {
"id": "280794135283124256",
"name": "Senior"
},
"start_date": "2012-01"
}
],
"education": [
{
"school": {
"id": "131182196916370",
"name": "Fatima School, Gonda"
},
"year": {
"id": "113125125403208",
"name": "2004"
},
"type": "High School"
}
],
"gender": "male"
}
我需要從上述JSON提取id
, name
, first_name
, last_name
, gender
。 以下是我編寫的程序,但是它以某種方式引發了異常。 我在做什么錯呢?
public class JSONParser {
private static final String URL = "https://graph.facebook.com/me?access_token=AAAG2HjMOAsEBAGBhjx2RqqLbOvnAZAxEPQ0X7ZC2JWY0YcQZDZDSSSAFTR";
private static HashMap<String, String> output = null;
public static void main(String[] args) throws Exception {
StringBuilder builder = new StringBuilder();
DefaultHttpClient httpclient = new DefaultHttpClient();
output = new HashMap<String, String>();
BufferedReader bufferedReader = null;
try {
HttpGet httpget = new HttpGet(URL);
httpget.getRequestLine();
HttpResponse response = httpclient.execute(httpget);
HttpEntity entity = response.getEntity();
//System.out.println(response.getStatusLine());
if (entity != null) {
InputStream inputStream = entity.getContent();
bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
for (String line = null; (line = bufferedReader.readLine()) != null;) {
builder.append(line).append("\n");
}
JSONObject jsonObject = new JSONObject(builder.toString());
JSONObject info = jsonObject.getJSONObject("id");
parseJSONObject(info, output);
}
} catch (Exception e) {
} finally {
try {
bufferedReader.close();
httpclient.getConnectionManager().shutdown();
} catch (IOException e) {
}
}
}
private static HashMap<String, String> parseJSONObject(JSONObject json,
HashMap<String, String> output) throws JSONException {
Iterator<String> keys = json.keys();
while (keys.hasNext()) {
String key = keys.next();
String val = null;
try {
JSONObject value = json.getJSONObject(key);
parseJSONObject(value, output);
} catch (Exception e) {
val = json.getString(key);
}
if (val != null) {
output.put(key, val);
}
}
return output;
}
}
例外:-
org.json.JSONException: JSONObject["id"] is not a JSONObject.
注意: -JSON是正確的。 在將其發布到這里之前,我對JSON做了一些修改。 因此,在此處復制粘貼可能會出錯。 但是假設JSON是正確的。仍然會引發異常。
您可以通過獲取字段的數據類型值來獲得字段,如下所示:
int id = jsonObject.getInt("id");
String name = jsonObject.getString("name");
String firstName = jsonObject.getString("first_name");
String lastName = jsonObject.getString("last_name");
String gender = jsonObject.getString("gender");
文件末尾有一個逗號結尾:
"gender": "male",
----------------^
您可以使用JSONLint來驗證您的JSON是否有效,然后再嘗試對其進行解析。 一些解析器可能有些松懈,可以原諒一些錯誤,但是大多數解析器都非常嚴格。
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