我试图为我的Android应用程序创建一个登录活动，我在对话框中说“尝试登录”。但该应用程序很快就会强行关闭。我会发布代码，任何帮助表示赞赏！
package com.fr31ght.etched;
import java.util.ArrayList;
import java.util.List;
import org.apache.http.NameValuePair;
import org.apache.http.message.BasicNameValuePair;
import org.json.JSONException;
import org.json.JSONObject;
import android.app.Activity;
import android.app.ProgressDialog;
import android.content.Intent;
import android.os.AsyncTask;
import android.os.Bundle;
import android.util.Log;
import android.view.View;
import android.view.View.OnClickListener;
import android.widget.Button;
import android.widget.EditText;
import android.widget.Toast;
public class LoginActivity extends Activity implements OnClickListener{
private EditText user, pass;
private Button Loginbtn;
//process dialog
private ProgressDialog pDialog;
//JSON parser class
JSONParser jsonParser = new JSONParser();
//php script location
private static final String LOGIN_URL =
"http://127.0.0.1//android- login/login.php";
//JSON element id's from php script
private static final String TAG_SUCCESS = "success";
private static final String TAG_MESSAGE = "message";
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_login);
//set up input fields
user = (EditText)findViewById(R.id.username);
pass = (EditText)findViewById(R.id.password);
//Set up buttons
Loginbtn = (Button)findViewById(R.id.loginbtn);
//Register Listeners
Loginbtn.setOnClickListener(this);
}
@Override
public void onClick(View v) {
// TODO Auto-generated method stub
switch (v.getId()){
case R.id.loginbtn:
new AttemptLogin().execute();
break;
default:
break;
}
}
class AttemptLogin extends AsyncTask<String, String, String>{
//Show progress dialog
boolean failure = false;
protected void onPreExecute(){
super.onPreExecute();
pDialog = new ProgressDialog(LoginActivity.this);
pDialog.setMessage("Attempting Login...");
pDialog.setIndeterminate(false);
pDialog.setCancelable(true);
pDialog.show();
}
@SuppressWarnings("deprecation")
@Override
protected String doInBackground(String... arg0) {
// TODO Auto-generated method stub
//check for success tag
int success;
String username = user.getText().toString();
String password = pass.getText().toString();
try{
//Building parmeters
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("username", username));
params.add(new BasicNameValuePair("password", password));
Log.d("request!", "Starting");
//Getting product details with HTTP REQUEST
JSONObject json = jsonParser.makeHttpRequest
(LOGIN_URL, "POST", params);
//check log for json response
Log.d("Login attempt", json.toString());
//Json Success tag
success = json.getInt(TAG_SUCCESS);
if (success == 1){
Log.d("Login Successful!", json.toString());
Intent i = new Intent(LoginActivity.this, Etched.class);
finish();
startActivity(i);
return json.getString(TAG_MESSAGE);
}
} catch (JSONException e){
e.printStackTrace();
}
return null;
}
protected void onPostExecute(String file_url){
//dismiss the dialog
pDialog.dismiss();
if(file_url !=null){
Toast.makeText(LoginActivity.this, file_url, Toast.LENGTH_LONG)
.show();
}
}
}
}
PHP代码
<?php
error_reporting(1);
//load and connect to MySQL database stuff
include('config.php');
if (!empty($_POST)) {
//gets user's info based off of a username.
$query = "SELECT id, username, password FROM users WHERE username
= :username";
$query_params = array(
':username' => $_POST['username']
);
try {
$stmt = $db->prepare($query);
$result = $stmt->execute($query_params);
}
catch (PDOException $ex) {
// For testing, you could use a die and message.
//die("Failed to run query: " . $ex->getMessage());
//or just use this use this one to product JSON data:
$response["success"] = 0;
$response["message"] = "Database Error1. Please Try Again!";
die(json_encode($response));
}
//This will be the variable to determine whether or not the
user's information is correct.
//we initialize it as false.
$validated_info = false;
//fetching all the rows from the query
$row = $stmt->fetch();
if ($row) {
//if we encrypted the password, we would unencrypt it here, but in
our case we just
//compare the two passwords
if ($_POST['password'] === $row['password']) {
$login_ok = true;
}
}
// If the user logged in successfully, then we send them to the private
members-only page
// Otherwise, we display a login failed message and show the login form
again
if ($login_ok) {
$response["success"] = 1;
$response["message"] = "Login successful!";
echo('successful!!');
die(json_encode($response));
} else {
$response["success"] = 0;
$response["message"] = "Invalid Credentials!";
echo('unsuccessful!!');
die(json_encode($response));
}
} else {
?>
<h1>Login</h1>
<form action="login.php" method="post">
Username:<br />
<input type="text" name="username" placeholder="username" />
<br /><br />
Password:<br />
<input type="password" name="password" placeholder="password"
value="" />
<br /><br />
<input type="submit" value="Login" />
</form>
<a href="register.php">Register</a>
<?php
}
?>
答案 0 :(得分：0)
在你的asynctask尝试这样做：
// Set connection
DefaultHttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(LOGIN_URL);
// Add parameters
List<NameValuePair> data = new ArrayList<NameValuePair>(2);
data.add(new BasicNameValuePair("username", username));
data.add(new BasicNameValuePair("password", password));
httppost.setEntity(new UrlEncodedFormEntity(data));
ResponseHandler<String> responseHandler = new BasicResponseHandler();
//Here you get the response from server
String response = httpclient.execute(httppost,responseHandler);
我在我的应用程序中使用它没有任何问题。