我正在尝试在 android 中进行会话处理过程。在这里，我已经通过 android 成功登录，现在我想处理登录用户的会话。这是我的 login_suer.java（android 部分）
package com.iwantnew.www;
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.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
public class login_user extends Activity{
private ProgressDialog pDialog;
JSONParser jsonParser = new JSONParser();
EditText login_email;
EditText login_password;
Button signin;
TextView error_msg;
private static String url_create_signin= "http://10.0.2.2/android_iwant/login_user.php";
// JSON Node names
private static final String TAG_SUCCESS = "success";
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.user_form);
// Edit Text
login_email = (EditText) findViewById(R.id.login_email);
login_password = (EditText) findViewById(R.id.login_password);
signin = (Button) findViewById(R.id.signin);
error_msg = (TextView) findViewById(R.id.error_msg);
signin.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
// creating new product in background thread
new CheckLogin().execute();
}
});
}
class CheckLogin extends AsyncTask<String, String, String> {
/**
* Before starting background thread Show Progress Dialog
* */
@Override
protected void onPreExecute() {
super.onPreExecute();
pDialog = new ProgressDialog(login_user.this);
pDialog.setMessage("Signing in..");
pDialog.setIndeterminate(false);
pDialog.setCancelable(true);
pDialog.show();
}
/**
* Creating product
* */
protected String doInBackground(String... args) {
//Building Parameters
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("email",login_email.getText().toString()));
params.add(new BasicNameValuePair("password", login_password.getText().toString()));
// getting JSON Object
// Note that create product url accepts POST method
JSONObject json = jsonParser.makeHttpRequest(url_create_signin,
"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 users
Intent i = new Intent(getApplicationContext(), post_item.class);
startActivity(i);
// closing this screen
finish();
} else {
// failed to sign in
error_msg.setText("Incorrect username/password");
}
} 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();
}
}
}
现在我需要在这个 java 文件中开始会话处理的想法。服务器端的代码如下：即login_user.php
<?php
session_start();
// array for JSON response
$response = array();
if(isset($_POST['email']) && isset($_POST['password'])){
$email = $_POST['email'];
$password = $_POST['password'];
// include db handler
require_once 'DB_Functions.php';
$db = new DB_Functions();
$user = $db->getUesrByEmailAndPassword($email, $password);
if ($user != false) {
// user found
// echo json with success = 1
$response["success"] = 1;
$response["uid"] = $user["unique_id"];
$response["user"]["name"] = $user["name"];
$response["user"]["email"] = $user["email"];
$response["user"]["created_at"] = $user["created_at"];
$response["user"]["updated_at"] = $user["updated_at"];
echo json_encode($response);
} else {
// user not found
// echo json with error = 1
$response["error"] = 1;
$response["error_msg"] = "Incorrect email or password!";
echo json_encode($response);
}
}
?>
上面这个 php 文件中使用的函数是 getUesrByEmailAndPassword($email, $password) 如下：
public function getUserByEmailAndPassword($email, $password) {
$result = mysql_query("SELECT * FROM users WHERE email = '$email'") or die(mysql_error());
// check for result
$no_of_rows = mysql_num_rows($result);
if ($no_of_rows > 0) {
$result = mysql_fetch_array($result);
$salt = $result['salt'];
$encrypted_password = $result['encrypted_password'];
$hash = $this->checkhashSSHA($salt, $password);
// check for password equality
if ($encrypted_password == $hash) {
// user authentication details are correct
//return $result;
session_start();
$_SESSION['clientId'] = $result[0];
$_SESSION['logged_in'] = TRUE;
}
} else {
// user not found
return false;
}
}
请帮助我使我的代码正常工作。任何帮助将不胜感激。任何包含此类问题解决方案的链接都可能对我有所帮助。谢谢你！