做Android的登录和注册时响应不包含任何数据 [英] Response does not contain any data when doing Android login and registration
问题描述
我正在开发一个Android手机应用程序登录和注册到MySQL，PHP和SQLite。我指的http://www.androidhive.info/2012/01/android-login-and-registration-with-php-mysql-and-sqlite/.不过，我面临着一些问题，因为PHP不断提醒我mysql_connect的德precation（）的。因此，我不得不切换所有MySQL的语法MySQLi的语法。并成功建立连接。当我通过在数据标签，姓名，电子邮件地址和密码的PHP文件，该文件是负责处理此数据并返回JSON响应，它显示的响应选项卡下：响应不包含任何数据。。
这是采用先进的REST客户端应用程序来测试反应的结果页面：
我的PHP文件，该文件要处理数据：
＆LT; PHP\r
\r
/ **\r
*文件来处理所有的API请求\r
*接受GET和POST\r
*\r
*每个请求将被标记进行标识\r
*反应将是JSON数据\r
\r
/ **\r
*检查POST请求\r
* /\r
如果（使用isset（$ _ POST ['标签']）及和放大器;！$ _ POST ['标签'] =''）{\r
//获取标签\r
$标签= $ _ POST ['标签'];\r
\r
//包括数据库处理程序\r
require_once'包括/ DB_Functions.php';\r
$ DB =新DB_Functions（）;\r
\r
//响应阵列\r
$响应=阵列（）;\r
\r
//检查标签类型\r
如果（$标记=='登录'）{\r
//请求的类型是检查登录\r
$电子邮件= $ _ POST [电子邮件];\r
$密码= $ _ POST ['密码'];\r
\r
//检查用户\r
$ USER = $ DB-GT＆; getUserByEmailAndPassword（$电子邮件，$密码）;\r
如果（$用户！= FALSE）{\r
//用户发现\r
$响应[错误] = FALSE;\r
$响应[UID] = $用户[UNIQUE_ID];\r
$响应[用户] [名称] = $用户[名];\r
$响应[用户] [电子邮件] = $用户[电子邮件];\r
$响应[用户] [created_at] = $用户[created_at];\r
$响应[用户] [的updated_at] = $用户[的updated_at];\r
回声json_en code（$响应）;\r
}其他{\r
//用户未找到\r
//回声JSON，错误= 1\r
$响应[错误] = TRUE;\r
$响应[ERROR_MSG] =！不正确的电子邮件或密码;\r
回声json_en code（$响应）;\r
}\r
}否则如果（$标记=='注册'）{\r
//请求的类型是注册新用户\r
$名称= $ _ POST ['名'];\r
$电子邮件= $ _ POST [电子邮件];\r
$密码= $ _ POST ['密码'];\r
\r
//检查用户是否已经存在\r
如果（$ DB-GT＆; isUserExisted（$电子邮件））{\r
//用户已经存在 - 错误响应\r
$响应[错误] = TRUE;\r
$响应[ERROR_MSG] =用户已经存在;\r
回声json_en code（$响应）;\r
}其他{\r
//存储用户\r
$ USER = $ DB-GT＆; storeUser（$名称，$电子邮件，$密码）;\r
如果（$用户）{\r
//用户成功存储\r
$响应[错误] = FALSE;\r
$响应[UID] = $用户[UNIQUE_ID];\r
$响应[用户] [名称] = $用户[名];\r
$响应[用户] [电子邮件] = $用户[电子邮件];\r
$响应[用户] [created_at] = $用户[created_at];\r
$响应[用户] [的updated_at] = $用户[的updated_at];\r
回声json_en code（$响应）;\r
}其他{\r
//用户未能保存\r
$响应[错误] = TRUE;\r
$响应[ERROR_MSG] =错误的发生Registartion;\r
回声json_en code（$响应）;\r
}\r
}\r
}其他{\r
//用户未能保存\r
$响应[错误] = TRUE;\r
$响应[ERROR_MSG] =不明'标签'的价值应该是'登录'或'注册';\r
回声json_en code（$响应）;\r
}\r
}其他{\r
$响应[错误] = TRUE;\r
$响应[ERROR_MSG] =所需的参数'标签'消失了！\r
回声json_en code（$响应）;\r
}\r
？＆GT; \r
有没有人有关于如何解决这个想法？
您回应阵列更改
$响应=阵列（标签=＆GT; $标记，错误=＆GT; FALSE）;
我刚刚换了同样来的mysqli和它的作品。
I am developing an Android mobile app login and registration to MySQL, PHP and SQLite. I am referring to http://www.androidhive.info/2012/01/android-login-and-registration-with-php-mysql-and-sqlite/. However I am facing some problem as the PHP keeps reminding me of the deprecation of mysql_connect(). So I had switch all the MySQL syntax to MySQLi syntax. And the connection is successfully established. When I pass in the data: "tag,name,email and password" The PHP file which is responsible for processing this data and return in JSON response, under the Response tab it's shown: Response does not contain any data..
This is the result page using Advanced Rest Client Application to test the response:
My php file which should process the data:
<?php
/**
* File to handle all API requests
* Accepts GET and POST
*
* Each request will be identified by TAG
* Response will be JSON data
/**
* check for POST request
*/
if (isset($_POST['tag']) && $_POST['tag'] != '') {
// get tag
$tag = $_POST['tag'];
// include db handler
require_once 'include/DB_Functions.php';
$db = new DB_Functions();
// response Array
$response = array();
// check for tag type
if ($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
$response["error"] = FALSE;
$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"] = TRUE;
$response["error_msg"] = "Incorrect email or password!";
echo json_encode($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"] = TRUE;
$response["error_msg"] = "User already existed";
echo json_encode($response);
} else {
// store user
$user = $db->storeUser($name, $email, $password);
if ($user) {
// user stored successfully
$response["error"] = FALSE;
$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"] = TRUE;
$response["error_msg"] = "Error occured in Registartion";
echo json_encode($response);
}
}
} else {
// user failed to store
$response["error"] = TRUE;
$response["error_msg"] = "Unknow 'tag' value. It should be either 'login' or 'register'";
echo json_encode($response);
}
} else {
$response["error"] = TRUE;
$response["error_msg"] = "Required parameter 'tag' is missing!";
echo json_encode($response);
}
?>
Does anyone have idea about how to solve this?
Change your response array to
$response = array("tag" => $tag, "error" => FALSE);
I've just changed the same to MYSQLI and it works.
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