Question

我尝试使用Android应用程序填充数据库表，但每当我点击提交按钮时都没有任何反应。它表示字符串不能转换为JSONObject。可能是什么原因？

这是错误

12-19 02:45:16.464: E/JSON Parser(1508): Error parsing data org.json.JSONException: Value <?xml of type java.lang.String cannot be converted to JSONObject

Java代码

// Edit Text
        inputName = (EditText) findViewById(R.id.inputName);
        inputPrice = (EditText) findViewById(R.id.inputPrice);
        //inputDesc = (EditText) findViewById(R.id.inputDesc);

        // Create button
        Button btnCreateProduct = (Button) findViewById(R.id.btnCreateProduct);

        // button click event
        btnCreateProduct.setOnClickListener(new View.OnClickListener() {

            @Override
            public void onClick(View view) {
                // creating new product in background thread
                new CreateNewProduct().execute();
            }
        });
    }

    /**
     * Background Async Task to Create new product
     * */
    class CreateNewProduct extends AsyncTask<String, String, String> {

        /**
         * Before starting background thread Show Progress Dialog
         * */
        @Override
        protected void onPreExecute() {
            super.onPreExecute();
            pDialog = new ProgressDialog(Leave.this);
            pDialog.setMessage("Creating Product..");
            pDialog.setIndeterminate(false);
            pDialog.setCancelable(true);
            pDialog.show();
        }

        /**
         * Creating product
         * */
        protected String doInBackground(String... args) {
            String name = inputName.getText().toString();
            String price = inputPrice.getText().toString();
            //String description = inputDesc.getText().toString();

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

            // getting JSON Object
            // Note that create product url accepts POST method
            JSONObject json = jsonParser.makeHttpRequest(url_create_product,
                    "POST", params);

            // check log cat fro response
            Log.d("Create Response", json.toString());

            // check for success tag
            try {
                int success = json.getInt(TAG_SUCCESS);

                if (success == 1) {
                    // successfully created product
                    //Toast.makeText(getApplicationContext(), "Successful", Toast.LENGTH_LONG).show();
                    //Toast.makeText(getApplicationContext(), "this is my Toast message!!! =)",
                              // Toast.LENGTH_LONG).show();

                    // closing this screen

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

            return null;
        }

        /**
         * After completing background task Dismiss the progress dialog
         * **/
        protected void onPostExecute(String file_url) {
            // dismiss the dialog once done
            pDialog.dismiss();
        }

    }
}

PHP

if (isset($_POST['reason'])) {

    $reason = $_POST['reason'];
    //$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
    $result = mysql_query("INSERT INTO leavelist(reason) VALUES('$reason')");

    // 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);

Answer 1

从错误消息中，似乎响应是xml。如果您尝试使用浏览器查看响应是否真的是JSON，或者您可以尝试在PHP文件中将返回类型设置为JSON，那将是很好的，如下所示：

header('Content-Type: application/json');

希望这会有所帮助。

试图从Android应用程序填充数据库？

1 个答案: