获取JSON异常没有价值吗? [英] Getting JSON Exception no value for?

问题描述

我正在尝试处理json响应，就像

I am trying to handle the json response, that is like

{"Status":true,"UserId":111,"FirstName":"dev","LastName":"dev","Gender":-1,"BirthdayDate":"0000-00-00","Phone":"","ProfilePicture":"","ProfilePicture60px":"","ProfilePicture120px":"","CountryId":-1,"Email":"droidwithmxxmail.com","Password":"******123","RegisterDate":"2015-05-08 20:08:07","SessionId":"fce248fe6499b7a9338a1b64554509eb77841"}

但得到org.json.JSONException: no value for exception

我的代码是这个.

if (jsonStr != null) {
            try {
                JSONObject jsonObj = new JSONObject(jsonStr);

            allres =jsonObj.getJSONArray(jsonStr);

                for (int i = 0; i < allres.length(); i++) {
                    JSONObject c = allres.getJSONObject(i);
                   userId = c.getString("UserId");
                }
            } catch (JSONException e) {
                e.printStackTrace();
            }

推荐答案

尝试以下代码:

if (jsonStr != null) {
  try {

    JSONObject jsonObj = new JSONObject(jsonStr);
   userId = jsonObj.getString("UserId");

  } catch (JSONException e) {
      e.printStackTrace();
}

您的JSON字符串中没有数组，因此从第一个JSONObject获取的对象已经是您的信息所在的地方.

Your JSON string doesn't have an array in it, so the object that you will be getting from the first JSONObject is already where your information lives.

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