android - 无法将数据发送到 Web 服务器的数据库

Question

当我通过浏览器直接访问PHP的链接时，它会在数据库中创建一个空记录。但是，当我使用应用程序发送数据时，它会失败。

Java：

protected String doInBackground(String... params) {
    String reg_url = "http://www.minigameserver.square7.ch/register.php";
    String user_name = params[0];
    String user_ac = params[1];
    String user_pw = params[2];
    try {
        URL url = new URL(reg_url);
        HttpURLConnection httpURLConnection = (HttpURLConnection) url.openConnection();
        httpURLConnection.setRequestMethod("POST");

        httpURLConnection.setDoOutput(true);
        OutputStream OS = httpURLConnection.getOutputStream();
        BufferedWriter bufferedWriter = new BufferedWriter(new OutputStreamWriter(OS,"UTF-8"));
        String data = URLEncoder.encode("user_name","UTF-8")+"="+URLEncoder.encode(user_name,"UTF-8")+"&"+
                URLEncoder.encode("user_ac","UTF-8")+"="+URLEncoder.encode(user_ac,"UTF-8")+"&"+
                URLEncoder.encode("user_pw","UTF-8")+"="+URLEncoder.encode(user_pw,"UTF-8");
        bufferedWriter.write(data);
        bufferedWriter.flush();
        bufferedWriter.close();
        OS.close();

        httpURLConnection.disconnect();

        return "done!";
    } catch (MalformedURLException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }
    return "fail!";
}

当我运行应用程序时，按摩“完成！”显示出来了。

所以，没有异常发生。但是，为什么没有新记录呢？

---

我看到有人用 teset 的用户名注册很累。他介意告诉谁做的，这样我就可以追踪这个帖子。

---

当我尝试 Arshak 提供的代码时。出现以下错误。 E/Surface：getSlotFromBufferLocked：未知缓冲区：0x7f7d9eaad530

注册.php：

<?php
require "db_connect.php";

$user_name = $_POST["user_name"];
$user_ac = $_POST["user_ac"];
$user_pw = $_POST["user_pw"];
$user_pw_enc = md5($user_pw);

$sql = "INSERT INTO user(user_name, user_ac, user_pw) VALUES ('$user_name', '$user_ac', '$user_pw_enc')";

$sql2 = "SELECT * FROM user WHERE user_name = '$user_name'";
$result2 = $conn->query($sql2);

$sql3 = "SELECT * FROM user WHERE user_ac = '$user_ac'";
$result3 = $conn->query($sql3);

if($result2->num_rows == 0 && $result3->num_rows == 0){
if ($conn->query($sql) === TRUE)
        echo "Your account is created successfully!\nYou can login now!";
else
    echo "Error: " . $sql . "<br>" . $conn->error;
}
else {
if($result2->num_rows > 0)  
    echo "The user name is used!<br>";
if($result3->num_rows > 0)
    echo "The user account is used!<br>";
}

?>

Answer 1

@Override
protected String doInBackground(String... params) {
    // Create a new HttpClient and Post Header
    HttpClient httpclient = new DefaultHttpClient();
    HttpPost httppost = new HttpPost("http://www.minigameserver.square7.ch/register.php");

    try {
       //add data
       List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(1);
       nameValuePairs.add(new BasicNameValuePair("user_name", params[0]));
       nameValuePairs.add(new BasicNameValuePair("user_ac", params[1]));
       nameValuePairs.add(new BasicNameValuePair("user_pw", params[2]));
       httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
       //execute http post
       HttpResponse response = httpclient.execute(httppost);

    } catch (ClientProtocolException e) {

    } catch (IOException e) {

    }
}

您可以使用 Volley Android 网络库发布您的数据。官方文档是https://developer.android.com/training/volley/simple.html。

Answer 2

我已经更新了你的代码，看看有没有数据被插入：

protected String doInBackground(String... params) {
String reg_url = "http://www.minigameserver.square7.ch/register.php";
String user_name = params[0];
String user_ac = params[1];
String user_pw = params[2];
try {
        URL url = new URL(reg_url);
        HttpURLConnection httpURLConnection = (HttpURLConnection) url.openConnection();
        httpURLConnection.setRequestMethod("POST");
        httpURLConnection.setDoOutput(true);
        httpURLConnection.setDoInput(true);
        httpURLConnection.connect();
        DataOutputStream OS = new DataOutputStream(httpURLConnection.getOutputStream());
        String data = URLEncoder.encode("user_name","UTF-8")+"="+URLEncoder.encode(user_name,"UTF-8")+"&"+
                URLEncoder.encode("user_ac","UTF-8")+"="+URLEncoder.encode(user_ac,"UTF-8")+"&"+
                URLEncoder.encode("user_pw","UTF-8")+"="+ URLEncoder.encode(user_pw,"UTF-8");
        OS .writeBytes(data);
        OS .flush();
        OS.close();

        int responseCode=httpURLConnection.getResponseCode();
        String messageFromServer;
        if (responseCode==200) {
            messageFromServer=fromInputStream(httpURLConnection.getInputStream());
            Log.v("RegistrationMessage ",messageFromServer);
            httpURLConnection.disconnect();
            return messageFromServer;
        }

        httpURLConnection.disconnect();

    } catch (IOException e) {
        e.printStackTrace();
    }
    return "fail!";
}

//Add this method inside your AsyncTask
public String fromInputStream(InputStream stream) throws IOException {
        InputStreamReader streamReader = new
                InputStreamReader(stream);
        BufferedReader bufferedReader = new BufferedReader(streamReader);
        StringBuilder builder = new StringBuilder();
        String tempString = "";
        while ((tempString = bufferedReader.readLine()) != null) {
            builder.append(tempString);
        }
        return builder.toString();
    }

Answer 3

我尝试测试该 URL，是的，它工作正常。由于已经有帖子显示对您的代码进行了一些修复，所以我只是给您一个提示。

首先，因为您不确定您可能会遇到什么异常。试着让它通用。删除一条catch() 语句，改为使用：

catch(Exception e){
    e.printStackTrace();
}

这样，所有异常都将被捕获，您可以轻松找到导致代码出现问题的原因。

接下来是，因为您正在与服务器通信，所以这将是一个请求-响应过程。实际上，您做的事情是正确的，因为在发送了一些请求后，我能够收到回复。但为了提高效率，您的服务器还应该响应布尔指示器之类的东西。假设它已成功注册将响应 1，否则将响应 0。然后您的移动应用程序将处理结果。

此外，您很可能无法收到该失败消息，因为它不在 try-catch 范围内。