Question

在我的Android中Jason Parsing之后我收到了错误

Caused by: java.lang.NullPointerException: Attempt to invoke virtual method 'java.lang.String org.json.JSONObject.toString()' on a null object reference

请帮助因为我是新的Php我无法弄清楚我的php响应错误或我的android代码错误

这是我的主要活动请帮助因为我是新的Php我无法弄清楚我的php响应错误或我的android代码错误

private ProgressDialog pDialog;

JSONParser jsonParser = new JSONParser();
EditText inputName;
EditText inputPrice;
EditText inputDesc;

// url to create new product
private static String url_create_product = "http://www.ledeveloper.in/create_product.php";

// JSON Node names
private static final String TAG_SUCCESS = "success";

String name,price,description;
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    // Edit Text
    inputName = (EditText) findViewById(R.id.inputName);
    inputPrice = (EditText) findViewById(R.id.inputPrice);
    inputDesc = (EditText) findViewById(R.id.inputDesc);

    Button btnCreateProduct = (Button) findViewById(R.id.btnCreateProduct);
    btnCreateProduct.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {
        name = inputName.getText().toString();
            price = inputPrice.getText().toString();
            description = inputDesc.getText().toString();

            new CreateNewProduct().execute();
        }
    });
}


class CreateNewProduct extends AsyncTask<String, String, String> {


    @Override
    protected void onPreExecute() {
        super.onPreExecute();
        pDialog = new ProgressDialog(MainActivity.this);
        pDialog.setMessage("Creating Product..");
        pDialog.setIndeterminate(false);
        pDialog.setCancelable(true);
        pDialog.show();
    }


    protected String doInBackground(String... args) {

        // Building Parameters
        List<NameValuePair> params = new ArrayList<NameValuePair>();
        params.add(new BasicNameValuePair("name", name));
        params.add(new BasicNameValuePair("price", price));
        params.add(new BasicNameValuePair("description", description));


        Log.e("abcd", String.valueOf(params));

        JSONObject json = jsonParser.makeHttpRequest(url_create_product,
                "POST", params);

        Log.d("Create Response", json.toString());

        try {
            int success = json.getInt(TAG_SUCCESS);

            if (success == 1) {
                // successfully created product
                Intent i = new Intent(getApplicationContext(), MainActivity.class);
                startActivity(i);

                // closing this screen
                finish();
            } else {
                // failed to create product
            }
        } catch (JSONException e) {
            e.printStackTrace();
        }

        return null;
    }
    protected void onPostExecute(String file_url) {
        // dismiss the dialog once done
        pDialog.dismiss();
    }
}

}

这是我的Json回复

{“success”：0，“message”：“缺少必填字段”}

这是我的PHP代码

$ response = array（）;

if (isset($_POST['name']) && isset($_POST['price']) && 
isset($_POST['description'])) {
$name = $_POST['name'];
$price = $_POST['price'];
$description = $_POST['description'];

// include db connect class
require_once __DIR__ . '/db_connect.php';

// connecting to db
$db = new DB_CONNECT();

// mysql inserting a new row
$result = mysql_query("INSERT INTO products(name, price, description) VALUES('$name', '$price', '$description')");

// check if row inserted or not
if ($result) {
    // successfully inserted into database
    $response["success"] = 1;
    $response["message"] = "Product successfully created.";

    // echoing JSON response
    echo json_encode($response);
} else {
    // failed to insert row
    $response["success"] = 0;
    $response["message"] = "Oops! An error occurred.";

    // echoing JSON response
    echo json_encode($response);
}
} else {
// required field is missing
$response["success"] = 0;
$response["message"] = "Required field(s) is missing";

// echoing JSON response
echo json_encode($response);
}
?>

这是我的Http请求文件公共类JSONParser {

static InputStream is = null;
static JSONObject jObj = null;
static String json = "";

// constructor
public JSONParser() {

}

// function get json from url
// by making HTTP POST or GET mehtod
public JSONObject makeHttpRequest(String url, String method,
                                  List<NameValuePair> params) {

    // Making HTTP request
    try {

        // check for request method
        if(method == "POST"){
            // request method is POST
            // defaultHttpClient
            DefaultHttpClient httpClient = new DefaultHttpClient();
            HttpPost httpPost = new HttpPost(url);
            httpPost.setEntity(new UrlEncodedFormEntity(params));

            HttpResponse httpResponse = httpClient.execute(httpPost);
            HttpEntity httpEntity = httpResponse.getEntity();
            is = httpEntity.getContent();

        }else if(method == "GET"){
            // request method is GET
            DefaultHttpClient httpClient = new DefaultHttpClient();
            String paramString = URLEncodedUtils.format(params, "utf-8");
            url += "?" + paramString;
            HttpGet httpGet = new HttpGet(url);

            HttpResponse httpResponse = httpClient.execute(httpGet);
            HttpEntity httpEntity = httpResponse.getEntity();
            is = httpEntity.getContent();
        }

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

    try {
        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");
        }
        is.close();
        json = sb.toString();
    } catch (Exception e) {
        Log.e("Buffer Error", "Error converting result " + e.toString());
    }

    // try parse the string to a JSON object
    try {
        jObj = new JSONObject(json);
    } catch (JSONException e) {
        Log.e("JSON Parser", "Error parsing data " + e.toString());
    }

    // return JSON String
    return jObj;

}

}

Answer 1

基本上，您不应该从doInBackground（）访问您的实例变量，因为它不是线程安全的。就像函数说的那样，它在一个单独的（后台）线程中运行。因此，你应该将名称，价格和描述作为参数传递给AsyncTask

试试这个

 class CreateNewProduct extends AsyncTask<String, String, String> {
    private String name,price,description;

    public CreateNewProduct(String name,String price, String desc){
    this.name=name;
        this.price=price;
        this.description=desc;
    }

    @Override
    protected void onPreExecute() {
        super.onPreExecute();
        pDialog = new ProgressDialog(MainActivity.this);
        pDialog.setMessage("Creating Product..");
        pDialog.setIndeterminate(false);
        pDialog.setCancelable(true);
        pDialog.show();
    }


    protected String doInBackground(String... args) {

        // Building Parameters
        List<NameValuePair> params = new ArrayList<NameValuePair>();
        params.add(new BasicNameValuePair("name", name));
        params.add(new BasicNameValuePair("price", price));
        params.add(new BasicNameValuePair("description", description));


        Log.e("abcd", String.valueOf(params));

        JSONObject json = jsonParser.makeHttpRequest(url_create_product,
                "POST", params);

        Log.d("Create Response", json.toString());

        try {
            int success = json.getInt(TAG_SUCCESS);

            if (success == 1) {
                // successfully created product
                Intent i = new Intent(getApplicationContext(), MainActivity.class);
                startActivity(i);

                // closing this screen
                finish();
            } else {
                // failed to create product
            }
        } catch (JSONException e) {
            e.printStackTrace();
        }

        return null;
    }
    protected void onPostExecute(String file_url) {
        // dismiss the dialog once done
        pDialog.dismiss();
    }
}

并将其称为

new CreateNewProduct(name,price,description).execute();

无法使用Php Json在数据库中发送数据

1 个答案: