java.lang.NullPointerException：尝试在空对象引用（帮助）上调用虚拟方法'java.lang.String java.lang.String.toString（）'[重复]

Question

当我按下骑手应用程序中的呼叫驱动程序按钮时。它在驱动程序应用程序上显示通知，但它崩溃了，因为它给出了调用空指针错误的错误。 在这一行出现空对象引用错误

(JSONObject jsonObject = new JSONObject(response.body().toString());)

这是我的 CustomerCall 课程

private void getDirection(double lat, double lng) {
    String requestAPI = null;
    try{

        requestAPI = "https://maps.googleapis.com/map/api/directions/json?"+
                "mode=driving&"+
                "transit_routing_preference=less_driving&"+
                "origin="+ Common.mLastLocation.getLatitude()+","+Common.mLastLocation.getLongitude()+"&"+
                "destination="+lat+","+lng+"&"+
                "key="+getResources().getString(R.string.google_direction_api);

        Log.d( "Sunny",requestAPI ); //print url for debug

        mService.getPath( requestAPI )
                .enqueue( new Callback<String>() {
                    @Override
                    public void onResponse(Call<String> call, Response<String> response) {

                            try {
                                JSONObject jsonObject = new JSONObject(response.body().toString());
                                JSONArray routes = jsonObject.getJSONArray("routes");

                                //after get routes,just get 1st element route
                                JSONObject object = routes.getJSONObject(0);

                                //after get 1st element, we need array with name "legs"
                                JSONArray legs = object.getJSONArray("legs");

                                //and get 1st element of legs array
                                JSONObject legsObject = legs.getJSONObject(0);

                                //Now get Distance
                                JSONObject distance = legsObject.getJSONObject("distance");
                                txtDistance.setText(distance.getString("text"));

                                //get time
                                JSONObject time = legsObject.getJSONObject("duration");
                                txtTime.setText(time.getString("text"));

                                //get address
                                String address = legsObject.getString("end_address");
                                txtAddress.setText(address);


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



                    @Override
                    public void onFailure(Call<String> call, Throwable t) {
                        Toast.makeText( CustomerCall.this, ""+t.getMessage(), Toast.LENGTH_SHORT ).show();

                    }
               } );

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

这里出现Logcat错误

E/AndroidRuntime: FATAL EXCEPTION: main
Process: com.example.shk.callingloader, PID: 12668
java.lang.NullPointerException: Attempt to invoke virtual method 'java.lang.String java.lang.String.toString()' on a null object reference
    at com.example.shk.callingloader.CustomerCall$1.onResponse(CustomerCall.java:90)
    at retrofit2.ExecutorCallAdapterFactory$ExecutorCallbackCall$1$1.run(ExecutorCallAdapterFactory.java:71)
    at android.os.Handler.handleCallback(Handler.java:790)
    at android.os.Handler.dispatchMessage(Handler.java:99)
    at android.os.Looper.loop(Looper.java:164)
    at android.app.ActivityThread.main(ActivityThread.java:7000)
    at java.lang.reflect.Method.invoke(Native Method)
    at com.android.internal.os.RuntimeInit$MethodAndArgsCaller.run(RuntimeInit.java:441)
    at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:1408)

Answer 1

错误信息说：

java.lang.NullPointerException: Attempt to invoke virtual method 'java.lang.String java.lang.String.toString()' on a null object reference
    at com.example.shk.callingloader.CustomerCall$1.onResponse(CustomerCall.java:90)

我不确定第 90 行是什么，但我会说 response.body() 是 null 并且您正在调用 toString() 。为您的方法添加一个空检查，它应该可以工作。

您可以将代码打包到 if 语句中：

if(response.body() != null){
    //your try-catch block
}

或者如果正文为空，您可以返回：

if(response.body() == null){
    return;
}

Answer 2

而不是

JSONObject jsonObject = new JSONObject(response.body().toString());

尝试使用

JSONObject jsonObject = new JSONObject(response.body()); // remove .tostring()