我正在尝试将我的Android应用程序连接到外部数据库。 我在ftp服务器上有一些php脚本，它为我的数据库提供了一些功能。 在我的phpmyadmin数据库中，我有一个表'用户'。 当我尝试插入一个新用户我的Android应用程序访问我的PHP脚本，但然后“mysql_query”返回总是假，用户不会出现在我的表在线数据库上。 在我的ftp站点中，我有一个index.php文件和一个文件夹“include”。在这个文件夹中我有三个带有函数和参数的文件：config.php，db_connect.php，db_function.php。
可能错误是愚蠢的，但我是php的新手。 谢谢大家。
这是我的java代码： 我使用registerUser使用getjsonfromurl
插入用户private static String loginURL = "php_script_adress_on_ftp";
private static String registerURL = "php_script_adress_on_ftp";
private static String login_tag = "login";
private static String register_tag = "register";
public JSONObject registerUser(String name, String email, String password){
// Building Parameters
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("tag", register_tag));
params.add(new BasicNameValuePair("name", name));
params.add(new BasicNameValuePair("email", email));
params.add(new BasicNameValuePair("password", password));
// getting JSON Object
JSONObject json = jsonParser.getJSONFromUrl(registerURL, params);
// return json
return json;
}
//这是jsonfromurl函数
public JSONObject getJSONFromUrl(String url, List<NameValuePair> params) {
// Making HTTP request
try {
// 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();
} 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();
Log.e("JSON", json);
} 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;
}
这是我从android app调用的index.php文件。 $ user = $ db-＆gt; storeUser（$ name，$ email，$ password）;返回始终为false，注册时出现错误编号“1”错误
<?php
/**
* File to handle all API requests
* Accepts GET and POST
*
* Each request will be identified by TAG
* Response will be JSON data
/**
* check for POST request
*/
if (isset($_POST['tag']) && $_POST['tag'] != '') {
// get tag
$tag = $_POST['tag'];
// include db handler
require_once 'include/DB_Functions.php';
$db = new DB_Functions();
// response Array
$response = array("tag" => $tag, "success" => 0, "error" => 0);
// check for tag type
if ($tag == 'login') {
// Request type is check Login
$email = $_POST['email'];
$password = $_POST['password'];
// check for user
$user = $db->getUserByEmailAndPassword($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);
}
} else
if ($tag == 'register') {
// Request type is Register new user
$name = $_POST['name'];
$email = $_POST['email'];
$password = $_POST['password'];
// check if user is already existed
if ($db->isUserExisted($email)) {
// user is already existed - error response
$response["error"] = 2;
$response["error_msg"] = "User already existed";
echo json_encode($response);
} else {
// store user
$user = $db->storeUser($name, $email, $password);
if ($user) {
// user stored successfully
$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 failed to store
$response["error"] = 1;
$response["error_msg"] = "Error occured in Registration";
echo json_encode($response);
}
}
} else {
echo "Invalid Request";
}
} else {
echo "Access Denied";
}
?>
这是我的config.php。以这种方式指定主机是否正确？我也试过http
<?php
/**
* Database config variables
*/
define("DB_HOST", "sql3.freemysqlhosting.net");
define("DB_USER", "my_user");
define("DB_PASSWORD", "my_psw");
define("DB_DATABASE", "mydb_name");
?>
这里是connect_db.php
<?php
class DB_Connect {
// constructor
function __construct() {
}
// destructor
function __destruct() {
// $this->close();
}
// Connecting to database
public function connect() {
require_once 'include/config.php';
// connecting to mysql
$con = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
// selecting database
mysql_select_db(DB_DATABASE);
// return database handler
return $con;
}
// Closing database connection
public function close() {
mysql_close();
}
}
?>
这里是db_functions.php
<?php
class DB_Functions {
private $db;
//put your code here
// constructor
function __construct() {
require_once 'DB_Connect.php';
// connecting to database
$this->db = new DB_Connect();
$this->db->connect();
}
// destructor
function __destruct() {
}
/**
* Storing new user
* returns user details
*/
public function storeUser($name, $email, $password) {
$uuid = uniqid('', true);
$hash = $this->hashSSHA($password);
$encrypted_password = $hash["encrypted"]; // encrypted password
$salt = $hash["salt"]; // salt
$result = mysql_query("INSERT INTO users(unique_id, name, email, encrypted_password, salt, created_at) VALUES('$uuid', '$name', '$email', '$encrypted_password', '$salt', NOW())");
// check for successful store
if ($result) {
// get user details
$uid = mysql_insert_id(); // last inserted id
$result = mysql_query("SELECT * FROM users WHERE uid = $uid");
// return user details
return mysql_fetch_array($result);
} else {
return false;
}
}
/**
* Get user by email and password
*/
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 $result;
}
} else {
// user not found
return false;
}
}
/**
* Check user is existed or not
*/
public function isUserExisted($email) {
$result = mysql_query("SELECT email from users WHERE email = '$email'");
$no_of_rows = mysql_num_rows($result);
if ($no_of_rows > 0) {
// user existed
return true;
} else {
// user not existed
return false;
}
}
/**
* Encrypting password
* @param password
* returns salt and encrypted password
*/
public function hashSSHA($password) {
$salt = sha1(rand());
$salt = substr($salt, 0, 10);
$encrypted = base64_encode(sha1($password . $salt, true) . $salt);
$hash = array("salt" => $salt, "encrypted" => $encrypted);
return $hash;
}
/**
* Decrypting password
* @param salt, password
* returns hash string
*/
public function checkhashSSHA($salt, $password) {
$hash = base64_encode(sha1($password . $salt, true) . $salt);
return $hash;
}
}
?>
编辑：我修改了db_function.php以返回mysql_error，但它什么也没有返回
public function storeUser($name, $email, $password,$error) {
$uuid = uniqid('', true);
$hash = $this->hashSSHA($password);
$encrypted_password = $hash["encrypted"]; // encrypted password
$salt = $hash["salt"]; // salt
$result = mysqli_query("INSERT INTO users(unique_id, name, email, encrypted_password, salt, created_at) VALUES('$uuid', '$name', '$email', '$encrypted_password', '$salt', NOW())");
// check for successful store
if (!$result) {
// get user details
$uid = mysqli_insert_id(); // last inserted id
$result = mysqli_query("SELECT * FROM users WHERE uid = $uid");
// return user details
$error = mysql_error();
return mysql_fetch_array($result);
} else {
$error = mysql_error();
return false;
}
}
我修改了index.php以返回mysql_error .... ....
$user = $db->storeUser($name, $email, $password,$error);
if ($user) {
// user stored successfully
$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 failed to store
$response["error"] = 1;
$response["error_msg"] = $error;
echo json_encode($response);
}
... ...
我也使用mysqli_query而不是mysql_query，但没有改变。
解： 我发现altervista重定向了我的HTTP POST，重定向后它变成了一个没有参数的HTTP GET。我不知道是什么原因。我有域名.com但它将请求重定向到.org。所以我直接向.org提出了请求，它开始工作了！
答案 0 :(得分：0)
解决方案：我发现altervista重定向了我的HTTP POST，重定向后它变成了一个没有参数的HTTP GET。我不知道是什么原因。我有域名.com但它将请求重定向到.org。所以我直接向.org提出了请求，它开始工作了！