[英]Unable to Send data in database Using Php Json
After Jason Parsing in my Android I am getting the error 在我的Android中进行Jason解析后,出现了错误
Caused by: java.lang.NullPointerException: Attempt to invoke virtual method 'java.lang.String org.json.JSONObject.toString()' on a null object reference
Please Help As I am New In Php I am unable to figure out that my php response is wrong or my android code is wrong 请帮忙,因为我是Php的新手,我无法弄清楚我的php响应错误或我的android代码错误
This Is My Main Activity Please Help As I am New In Php I am unable to figure out that my php response is wrong or my android code is wrong 这是我的主要活动,请帮忙,因为我是Php的新手,我无法找出我的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();
}
}
} }
This is My Json Response 这是我的杰森回应
{"success":0,"message":"Required field(s) is missing"} {“成功”:0,“消息”:“必填字段丢失”}
This Is my php code 这是我的PHP代码
$response = array(); $ 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);
}
?>
This Is My Http Request file public class JSONParser { 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;
}
}
Basically, you shouldn't access your instance variables from doInBackground() because it's not thread-safe. 基本上,您不应该从doInBackground()访问实例变量,因为它不是线程安全的。 Like the function says, it runs in a separate (background) thread.So, you should pass name, price and description as a parameter to the AsyncTask
就像函数所说的那样,它运行在一个单独的(后台)线程中,因此,您应该将名称,价格和描述作为参数传递给AsyncTask
Try this 尝试这个
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();
}
}
and call this like 像这样称呼
new CreateNewProduct(name,price,description).execute();
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