Question

我想将数据发布到url并获取Null pointer exception 我的JSON URL包含

{
  "Details":
    [
      {
        "Status":"NO UPDATES"
      }
    ]
}

我收到了错误行：

String status = object.getString("Status").trim(); //error Line

完整代码：

btnPost = (Button)findViewById(R.id.btnPost);

    btnPost.setOnClickListener(this);
    btnPost.setOnClickListener(new OnClickListener()
    {
        @SuppressWarnings("null")
        @Override
        public void onClick(View arg0)
        {
            try
            {
                String postReceiverUrl = "http://";
                Log.v(TAG, "postURL: " + postReceiverUrl);
                // HttpClient
                @SuppressWarnings("resource")
                HttpClient httpClient = new DefaultHttpClient();
                // post header
                HttpPost httpPost = new HttpPost(postReceiverUrl);

                jsonobject.put("IDNo", IDNo.getText().toString());
                jsonobject.put("Position", Position.getText().toString());
                jsonobject.put("Data", Data.getText().toString());

                HttpResponse response = httpClient.execute(httpPost);

                String jsonResult = inputStreamToString(response.getEntity().getContent()).toString();

                Log.v("jsonResult",jsonResult);

                JSONObject object = new JSONObject(jsonResult);

                String status = object.getString("Status").trim();

                Toast.makeText(getBaseContext(), "Please wait...",100).show();

                if(status.toString().equals("SUCCESS"))
                {
                    Intent i = new Intent(LoginPage.this,MainActivity.class);
                    i.setFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);
                    startActivity(i);
                }
                if(status.toString().equals("FAILED"))
                {
                    Toast.makeText(getBaseContext(), "Wrong Credentials",100).show();
                }

                else
                {
                    Toast.makeText(getBaseContext(), "Details Inserted",100).show();
                }

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

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


        }

    });

Answer 1

正确解析JSONObject object = new JSONObject(jsonResult); JSONArray array=object.getJsonarray("Details"); for(int i=0;i<array.length();i++) { JSONObject innerObject=array.getJsonObject(i); String s= innerObject.getString("Status"); }，你会看到你的回复： -

JSON

首先尝试制作{ "Details":[{"Status":"NO UPDATES"}]}的对象而不是JSONObject之后的对象，请看下面的示例： -

JSONArray

Answer 2

Modift String status = object.getString("Status").trim();到

String status = object.get("Details").getAsJsonArray()[0].getString("Status").trim();

Answer 3

您正在使用getString("Status")，如果密钥在JSON中不可用，则可能会返回null。我建议你使用optString("Status")如果密钥不可用，它将返回空白字符串。

JSONObject jsonObject = new JSONObject(stringJSON);
JSONArray jsonArray = jsonObject.optJSONArray("Details");
if(jsonArray != null){
   for(int i = 0; i < jsonArray.length(); i++){
     JSONObject jObject = jsonArray.optJSONObject(i);
     if(jObject != null){
       String strStatus = jObject.optString("Status");
     }
   }
}

Answer 4

试用此代码

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.2.0/jquery.min.js"></script>
<div>abc</div>
<button type="button" id="zoom-in">zoom in</button>
<button type="button" id="zoom-out">zoom out</button>
<script>
    $('#zoom-in').click(function() {
        updateZoom(0.1);
    });

    $('#zoom-out').click(function() {
        updateZoom(-0.1);
    });


    zoomLevel = 1;

    var updateZoom = function(zoom) {
        zoomLevel += zoom;
        $('body').css({
            zoom: zoomLevel,
            '-moz-transform': 'scale(' + zoomLevel + ')'
        });
    }
</script>

Answer 5

你的代码应该是这样的

if j

显示空指针异常的JSON对象

5 个答案: