Java 错误解析数据 org.json.JSONException: End of input at character 0 of of
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Error parsing data org.json.JSONException: End of input at character 0 of
提问by ali
I'm trying to parse JSON data which is coming from this url.
我正在尝试解析来自这个 url 的JSON 数据。
But I am getting these errors:
但我收到这些错误：
03-27 16:48:21.019: E/Buffer Error(23717): Error converting result java.lang.NullPointerException
03-27 16:48:21.059: E/JSON Parser(23717): Error parsing data org.json.JSONException: End of input at character 0 of
03-27 16:48:21.019：E/缓冲区错误（23717）：错误转换结果 java.lang.NullPointerException
03-27 16:48:21.059：E/JSON 解析器（23717）：解析数据时出错 org.json.JSONException：在字符 0 处输入结束
Wwhen I debug my code; getJsonFromUrl()method returns null jobject. Here is the JSONParser classwhich I used. What's causing the error?
当我调试我的代码时；getJsonFromUrl()方法返回null jobject。这是我使用的JSONParser 类。导致错误的原因是什么？
public class JSONParser {
static InputStream iStream = null;
static JSONArray jarray = null;
static JSONObject jObj= null;
static String json = "";
public JSONParser() {
}
public JSONObject getJSONFromUrl(String url) {
// Making HTTP request
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
try {
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
InputStream is = httpEntity.getContent();
BufferedReader reader = new BufferedReader(new InputStreamReader(is, "iso-8859-1"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
iStream.close();
json = sb.toString();
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// try parsing the string to a JSON object
try {
if (json != null) {
jObj = new JSONObject(json);
} else {
jObj = null;
}
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
// return JSON String
return jObj;
}
}
I'm making the call for this method from another class using these lines. (url parameter is defined at top)
我正在使用这些行从另一个类调用此方法。（url 参数定义在顶部）
JSONParser jParser = new JSONParser();
final JSONObject jobject = jParser.getJSONFromUrl(url);
采纳答案by vzamanillo
You are trying to get the JSON content using a HTTP POST method instead of the appropiated GET method (W3schools.com GET vs.POST), modify your source code to simplify and fix your HTTP request
您正在尝试使用 HTTP POST 方法而不是适当的 GET 方法（W3schools.com GET vs.POST）获取 JSON 内容，请修改您的源代码以简化和修复您的 HTTP 请求
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpGet get = new HttpGet(url);
try {
HttpResponse httpResponse = httpClient.execute(get);
String json = EntityUtils.toString(httpResponse.getEntity());
System.out.println(json);
....
....
} catch (Exception e) {
....
}
