Android HTTPPOST执行两次PHP脚本

Question

我正在开发一个基本的Android APK，它将成为我运行的网站的基本移动版本。 APK的一部分使人们可以注册用户帐户（基本注册）。 我正在使用HTTPPOST将用户表单信息发送到PHP脚本。 我所有的错误检查都能正常工作（密码匹配，长度等），但是从有效提交中得到的响应中，我可以告诉它尝试两次运行PHP。 APK会收到一条错误消息，并告诉用户已使用了用户名和密码，但这是因为php获得了运行用户名和密码，然后插入了数据，然后由于某种原因再次运行了，并且报告了错误捕获，而不仅仅是在首次运行时报告成功。 如果取出错误捕获以查找重复的用户名和电子邮件，则可以在数据库中看到两个插入。 为什么PHP页面两次运行？

脚步：

1. 清除用户数据库。
2. 插入有效的注册信息（以免出错）。

结果

APK收到显示错误消息，提示已使用用户名和密码。 SQL数据库显示已插入表单中的信息。

档案

我将登录记录在APK中，似乎只调用一次HTTPPOST。 以下是JAVA文件和PHP页面。 任何煽动将不胜感激。

JAVA文件

Register.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            Log.d("Registration: ", "Button Pressed");
            dialog = ProgressDialog.show(Register.this, "",
                    "Creating account...", true);
            new Thread(new Runnable() {
                public void run() {
                    new RegisterUser().execute("");
                }
            }).start();
        }
    });
}

private class RegisterUser extends AsyncTask<String, Void, String> {

    @Override
    protected String doInBackground(String... arg0) {
        try {


            httpclient = new DefaultHttpClient();
            // This is the page that will get all the information to create the account
            HTTPPOST = new HttpPost("http://www.linetomyphppage.php");

            NAMEVALUEPAIRS = new ArrayList<NameValuePair>();

            NAMEVALUEPAIRS.add(new BasicNameValuePair("username",username.getText().toString().trim()));  // $Edittext_value = $_POST['Edittext_value'];
            NAMEVALUEPAIRS.add(new BasicNameValuePair("userEmail",emailaddress.getText().toString().trim()));
            NAMEVALUEPAIRS.add(new BasicNameValuePair("password1",password1.getText().toString().trim()));
            NAMEVALUEPAIRS.add(new BasicNameValuePair("password2",password2.getText().toString().trim()));
            NAMEVALUEPAIRS.add(new BasicNameValuePair("newsletter",subscribe));

            HTTPPOST.setEntity(new UrlEncodedFormEntity(NAMEVALUEPAIRS));
            HTTPRESPONSE = httpclient.execute(HTTPPOST);
            ResponseHandler<String> responseHandler = new BasicResponseHandler();
            final String response = httpclient.execute(HTTPPOST, responseHandler);
            Log.d("Registration: ", "Executing Post");
            errorcode = response;

            // Redirect the user depending on the response
            if (response.equalsIgnoreCase("success")) {
               // Toast.makeText(Register.this, "Registration successful", Toast.LENGTH_SHORT).show();
                Log.d("Registration: ", "Registration successful");
                valid = "Successful";
            } else {
                Log.d("Registration: ", "Registration failed");
                valid = "Invalid";
              //  Toast.makeText(Register.this, "Registration failed: " + response, Toast.LENGTH_SHORT).show();
            }


        } catch (Exception e) {
            System.out.println("Exception here : " + e);
        }
        dialog.dismiss();
        return "Executed";
    }
    @Override
    protected void onPostExecute(String result){

        if(valid.equalsIgnoreCase("Invalid")){
            Log.d("Registration: ", "Error generated");
            String code = errorcode.replace("Invalid","");
            Toast.makeText(Register.this, "Registration failed: " + code, Toast.LENGTH_SHORT).show();
        }else{
            Toast.makeText(Register.this, "Registration successful", Toast.LENGTH_SHORT).show();
            startActivity(new Intent(Register.this, HomeScreen.class));
            finish();
        }
    }
}

PHP方面

<?php 
// Connecting to database
$connect = $_SERVER['DOCUMENT_ROOT'];
$connect .= "pathtoconnect.php";
include_once($connect);

// Get all the variables being passed from the APK
$username = mysql_real_escape_string($_POST['username']);
$password1 = mysql_real_escape_string($_POST['password1']);
$password2 = mysql_real_escape_string($_POST['password2']);
$email = mysql_real_escape_string($_POST['userEmail']);
$newsletter = mysql_real_escape_string($_POST['newsletter']);
$pass = 1;
$error = false;

// Check if the username is valid. 
if (preg_match('/[^0-9a-z-_]/i', $username) == 1) {
    $error = 'You can not use spaces or strange characters in a usermame.';
    $pass = 0;      
}
if(strlen($username) < 5) { // Ensure the length is at least 5
    $error = 'Username must be at least 5 characters.';
    $pass= 0 ;
}
if(strlen($password1) < 5) { // Ensure the length is at least 5
    $error = 'Passwords have to be at least slightly challenging (5+ characters).';
    $pass= 0 ;
}

if($password1 != $password2) { // Check for passwords to match
    $error = "Your passwords did not match. Please try again.";
    $pass= 0 ;  
}

if (!filter_var($email, FILTER_VALIDATE_EMAIL)) { // Check to make sure email address is valid
    $error = "The email address you entered is not valid. Double check it.";
    $pass= 0 ;
}

// check if username or email is already in use
$checking = mysql_query("SELECT COUNT(user_id) AS total FROM users WHERE username='$username' OR email='$email'");
$checking = mysql_fetch_array($checking);
if($checking['total'] > 0) {
    $error = 'The username or email you are trying to use is already in use. Please go to the website if you forgot your password to reset it.';    
    $pass= 0 ;
}
if(!$error) {
    echo 'success';
    $time = time();
    mysql_query("INSERT INTO users (username, password, email, date_created, subscription) VALUES ('$username', md5('$password1'), '$email',  '$time', '$subscribe')"); 


    }
else {
    echo 'Invalid'.$error;  
}

？>

该设备正在报告重复错误，并且正如我提到的那样，数据已插入，因此不会像“成功”那样被回显。 以下是JAVA文件中的日志，显示了该日志仅得到失败的响应。

02-04 22:31:37.442  26159-26159/com.demo D/Registration:﹕ Button Pressed
02-04 22:31:37.862  26159-26559/com.demo D/Registration:﹕ Executing Post
02-04 22:31:37.862  26159-26559/com.demo D/Registration:﹕ Registration failed
02-04 22:31:37.872  26159-26159/com.demo D/Registration:﹕ Error generated

我建立了一个基本的html表单来传递数据，该方法效果很好，一次执行了php，所以我很茫然。 Android / JAVA对我来说是新手，所以这是实现此目的的错误途径吗？

谢谢

Answer 1

您两次调用“ httpclient.execute（..）”：

HTTPRESPONSE = httpclient.execute(HTTPPOST);
ResponseHandler<String> responseHandler = new BasicResponseHandler();
final String response = httpclient.execute(HTTPPOST, responseHandler);

删除其中之一应该可以解决您的问题:)

Answer 2

您不需要将AsynTask子类（RegisterUser）调用到线程中，因为它是一个线程itelt，它耗尽了主要的IU线程。 只需调用execute方法。

而不是做：

new Thread(new Runnable() {
            public void run() {
                new RegisterUser().execute("");
            }
        }).start();

做：

                new RegisterUser().execute("");

可能可以解决问题。

此外，您将从DefaultHttpClient对象的第一行和第三行调用两次execute方法：

HTTPRESPONSE = httpclient.execute(HTTPPOST);
ResponseHandler<String> responseHandler = new BasicResponseHandler();
final String response = httpclient.execute(HTTPPOST, responseHandler);

Answer 3

它不是一个真正的答案，但是我将代码转换为使用JSON请求而不是httppost方法，并且效果更好。

Log.d("JSON", "Running the JSON script");
            JSONObject json = jsonParser.makeHttpRequest(url_register, "POST", params);

            //Getting specific Response
            try {
                RESPONSE_STATUS = json.getInt("success");
                RESPONSE_MESSAGE = json.getString("message");

            } catch (JSONException e) {
                RESPONSE_MESSAGE = "An error has occured during the registration";
                e.printStackTrace();
            }
            Log.d("JSON", json.toString());
            Log.d("Registration: ", "Response Status: " + RESPONSE_STATUS);
            Log.d("Registration: ", "Response: " + RESPONSE_MESSAGE);

            // Redirect the user depending on the response
            if (RESPONSE_STATUS == 1) {
                Log.d("Registration: ", "Registration successful");
                valid = "Successful";
            } else {
                Log.d("Registration: ", "Registration failed");
                valid = "Invalid";
            }

确实需要完全重写php才能将响应代码更改为数组。

同样，这不能解决我的原始问题，但是是一种可行的替代方法。