通过chrome，我可以使用数据库(MySQL)的成员登录，我可以创建一个新用户。当我使用应用程序时，我可以创建一个新用户，但是应用程序崩溃了。
使用简单的php(如下所示)，我得到了一个正确的响应，并启动了新的活动。这里的问题是正确和不正确的凭据都可以工作。
<?php
$response = array();
if (isset($_POST['username'], $_POST['password'])) {
$response["success"] = 1;
$response["message"] = "User Name and password are set";
} else {
$response["success"] = 0;
$response["message"] = "User Name and password not set";
}
echo json_encode($response);
下面是我的login.java文件:
package com.amity.paul.amity;
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.content.SharedPreferences;
import android.os.AsyncTask;
import android.os.Bundle;
import android.preference.PreferenceManager;
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;
import library.JSONParser;
public class login extends Activity implements OnClickListener{
private EditText user, pass;
private Button mSubmit, mRegister;
// Progress Dialog
private ProgressDialog pDialog;
// JSON parser class
JSONParser jsonParser = new JSONParser();
//php login script location:
//localhost :
//testing on your device
//put your local ip instead, on windows, run CMD > ipconfig
//or in mac's terminal type ifconfig and look for the ip under en0 or en1
// private static final String LOGIN_URL = "http://xxx.xxx.x.x:1234/webservice/login.php";
//testing on Emulator:
private static final String LOGIN_URL = "http://10.0.2.2/amity/login.php";
//testing from a real server:
//private static final String LOGIN_URL = "http://www.yourdomain.com/webservice/login.php";
//JSON element ids from repsonse of php script:
private static final String TAG_SUCCESS = "success";
private static final String TAG_MESSAGE = "message";
@Override
protected void onCreate(Bundle savedInstanceState) {
// TODO Auto-generated method stub
super.onCreate(savedInstanceState);
setContentView(R.layout.login);
//setup input fields
user = (EditText)findViewById(R.id.username);
pass = (EditText)findViewById(R.id.password);
//setup buttons
mSubmit = (Button)findViewById(R.id.login);
mRegister = (Button)findViewById(R.id.register);
//register listeners
mSubmit.setOnClickListener(this);
mRegister.setOnClickListener(this);
}
@Override
public void onClick(View v) {
// TODO Auto-generated method stub
switch (v.getId()) {
case R.id.login:
new AttemptLogin().execute();
break;
case R.id.register:
Intent i = new Intent(this, register.class);
startActivity(i);
break;
default:
break;
}
}
class AttemptLogin extends AsyncTask<String, String, String> {
/**
* Before starting background thread Show Progress Dialog
* */
boolean failure = false;
@Override
protected void onPreExecute() {
super.onPreExecute();
pDialog = new ProgressDialog(login.this);
pDialog.setMessage("Attempting login...");
pDialog.setIndeterminate(false);
pDialog.setCancelable(true);
pDialog.show();
}
@Override
protected String doInBackground(String... args) {
// TODO Auto-generated method stub
// Check for success tag
int success;
String username = user.getText().toString();
String password = pass.getText().toString();
try {
// Building Parameters
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 by making HTTP request
JSONObject json = jsonParser.makeHttpRequest(
LOGIN_URL, "POST", params);
// check your 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());
//save user data
SharedPreferences sp = PreferenceManager
.getDefaultSharedPreferences(login.this);
SharedPreferences.Editor edit = sp.edit();
edit.putString("username",username);
edit.commit();
Intent i = new Intent(login.this, home.class);
finish();
startActivity(i);
return json.getString(TAG_MESSAGE);
}else{
Log.d("Login Failure!", json.getString(TAG_MESSAGE));
return json.getString(TAG_MESSAGE);
}
} 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 product deleted
pDialog.dismiss();
if (file_url != null){
Toast.makeText(login.this, file_url, Toast.LENGTH_LONG).show();
}
}
}
}
然后我把php改成更高级的，如下所示:
<?php
$response = array();
if (isset($_POST['username'], $_POST['password'])) {
require("config.inc.php");
try {
$query = "
SELECT
id,
username,
password
FROM users
WHERE
username = :username
AND password = :password
";
$stmt = $db->prepare($query);
$result = $stmt->execute(array(':username' => $_POST['username'],
':password' => $_POST['password']
));
if($stmt->rowCount() == 1){
$response["success"] = true;
$response["message"] = "user found";
}else{
$response["success"] = false;
$response["message"] = "user not found";
}
}
catch (PDOException $ex) {
$response["success"] = false;
$response["message"] = "Database Error1. Please Try Again!";
}
} else {
$response["success"] = false;
$response["message"] = "User Name and password not set";
}
echo json_encode($response);
问题是解析错误。Logcat:
08-16 15:39:40.620 841-856/com.amity.paul.amity E/JSON Parser﹕ Error parsing data org.json.JSONException: Value of type java.lang.String cannot be converted to JSONObject
08-16 15:39:40.630 841-856/com.amity.paul.amity W/dalvikvm﹕ threadid=11: thread exiting with uncaught exception (group=0xb3a21ba8)
08-16 15:39:40.670 841-856/com.amity.paul.amity E/AndroidRuntime﹕ FATAL EXCEPTION: AsyncTask #1
Process: com.amity.paul.amity, PID: 841
java.lang.RuntimeException: An error occured while executing doInBackground()
at android.os.AsyncTask$3.done(AsyncTask.java:300)
at java.util.concurrent.FutureTask.finishCompletion(FutureTask.java:355)
at java.util.concurrent.FutureTask.setException(FutureTask.java:222)
at java.util.concurrent.FutureTask.run(FutureTask.java:242)
at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:231)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1112)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:587)
at java.lang.Thread.run(Thread.java:841)
Caused by: java.lang.NullPointerException
at com.amity.paul.amity.login$AttemptLogin.doInBackground(login.java:117)
at com.amity.paul.amity.login$AttemptLogin.doInBackground(login.java:97)
at android.os.AsyncTask$2.call(AsyncTask.java:288)
at java.util.concurrent.FutureTask.run(FutureTask.java:237)
at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:231)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1112)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:587)
at java.lang.Thread.run(Thread.java:841)
请使解决方案"新手友好"，因为我是新的android应用程序开发。谢谢!
编辑:我使用以下php修复了这个问题:
<?php
$response = array();
if (isset($_POST['username'], $_POST['password'])) {
require("config.inc.php");
try {
$query = "
SELECT
id,
username,
password
FROM users
WHERE
username = :username
AND password = :password
";
$stmt = $db->prepare($query);
$result = $stmt->execute(array(':username' => $_POST['username'],
':password' => $_POST['password']
));
if($stmt->rowCount() == 1){
$response["success"] = true;
$response["message"] = "user found";
}else{
$response["success"] = false;
$response["message"] = "user not found";
}
}
catch (PDOException $ex) {
$response["success"] = false;
$response["message"] = "Database Error1. Please Try Again!";
}
} else {
$response["success"] = false;
$response["message"] = "User Name and password not set";
}
echo json_encode($response);
最后一个问题，在logcat中，它会告诉我success = true/false。我如何使消息显示不正确的凭据，如果输入了正确的组合，启动新的活动?
你应该删除数组中的':'字符你传递给prepare():
$result = $stmt->execute(array('username' => $_POST['username'],
'password' => $_POST['password']
因为否则PDO会返回一个空响应，或者更糟的是，一个"警告"或"错误"php语句，这些语句将在最后生成的JSON之前加起来，从而生成一个解析异常。您的php响应看起来像:
Error in line BLA BLA BLA ...
["success": "false", "message":"blablabla"]
你将无法用你的Java应用程序解析。
所以它实际上不是一个问题在你的Android代码!
请求中的':'表示"在给定数组中检查':'之后的值"。
假设第二个php代码是用于登录的，我之前没有使用isset函数，但您可以使用此代码。这真的很简单。
if (!empty($_POST)) {
//gets user's info based off of a username.
$query = "
SELECT
username,
password
FROM login
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;
$response->{"message"} = "Database Error";
die(json_encode($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();
$login_ok = false;
if ($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;
$response->{"message"} = "Login Succesfull";
die(json_encode($response));
} else {
$response->{"success"} = 0;
$response->{"message"} = "Wrong Password";
die(json_encode($response));
}
}