在安卓上需要帮助。一直有这个错误Error Parsing data org.json.JSONException: Value <html> of type java.lang.String cannot be converted to JSONObject
下面是我运行它时给我错误的页面。
public class Login extends Activity {
//URL to get User Data
private static String URL_GET = "http://10.0.2.2/android_connect/get_authentication.php";
// JSON Node names
private static final String TAG_SUCCESS = "success";
private static final String TAG_USERS = "Users";
private static final String TAG_EMAIL = "Email";
private static final String TAG_PASSWORD = "Password";
JSONParser jParser = new JSONParser();
String dbPassword = null;
// users JSONArray
JSONArray Users = null;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
// setting default screen to login.xml
setContentView(R.layout.login);
Button btn = (Button) findViewById(R.id.btnLogin);
// Listening to register new account link
btn.setOnClickListener(new View.OnClickListener() {
public void onClick(View v) {
String password = ((EditText) findViewById(R.id.password)).getText().toString();
new getAuthentication().execute();
if (password.equals(dbPassword)) {
// Switching to Main screen
Intent i = new Intent(getApplicationContext(), EchoSphere.class);
startActivity(i);
}
}
});
}
class getAuthentication extends AsyncTask<String, String, String>{
@Override
protected String doInBackground(String... params) {
runOnUiThread(new Runnable() {
public void run() {
// Check for success tag
int success;
try {
String email = ((EditText) findViewById(R.id.email)).getText().toString();
Log.d("email:", email);
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair(TAG_EMAIL, email));
JSONObject json = jParser.makeHttpRequest(URL_GET, "GET", params);
Log.d("Password:", json.toString());
success = json.getInt(TAG_SUCCESS);
if (success == 1) {
Users = json.getJSONArray(TAG_USERS);
JSONObject c = Users.getJSONObject(0);
dbPassword = c.getString(TAG_PASSWORD);
Log.d("DBPW:", dbPassword);
}
} catch (JSONException e) {
e.printStackTrace();
}
}
});
return null;
}
}
}
这是 PHP 脚本，我已经验证了 PHP，它显示可以正常工作。
<?php
// array for JSON response
$response = array();
// include db connect class
require_once __DIR__ . '/db_connect.php';
// connecting to db
$db = new DB_CONNECT();
// check for post data
if (isset($_GET["Email"])) {
$email = $_GET['Email'];
// get a product from products table
$result = mysql_query("SELECT Email, Password FROM Users WHERE Email = '$email'");
if (!empty($result)) {
// check for empty result
if (mysql_num_rows($result) > 0) {
$result = mysql_fetch_array($result);
$user[] = array();
$user["Email"] = $result["Email"];
$user["Password"] = $result["Password"];
// user node
$response["Users"] = array();
array_push($response["Users"], $user);
// success
$response["success"] = 1;
// echoing JSON response
echo json_encode($response);
} else {
// no product found
$response["success"] = 0;
$response["message"] = "No User found";
// echo no users JSON
echo json_encode($response);
}
} else {
// no product found
$response["success"] = 0;
$response["message"] = "No User found";
// echo no users JSON
echo json_encode($response);
}
} else {
// required field is missing
$response["success"] = 0;
$response["message"] = "Required field(s) is missing";
// echoing JSON response
echo json_encode($response);
}
?>
结果如图
{"Users":[{"0":[],"Email":"1","Password":"123"}],"success":1}
我不知道错误在哪里。