Question

当我直接通过浏览器访问PHP链接时，它会在数据库中创建一个空记录。但是，当我使用app发送数据时，它会失败。

爪哇：

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，按摩完成了！＆＃34;显示。

所以，没有异常发生。但是，为什么没有新记录？

我看到有人厌倦了使用teset的用户名注册。他是否想知道是谁这样做我可以追查这个帖子。

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

register.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

我已更新您的代码，看看数据是否已插入：

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 2

@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 Networking Library发布您的数据。官方文件是https://developer.android.com/training/volley/simple.html。

Answer 3

我尝试测试网址，是的，它正在运行。由于已经有帖子显示您的代码有一些修复，我只是给你一个提示。

首先，因为您不确定可能会出现什么异常。尽量使它一般。删除一个catch()语句，然后使用：

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

通过这种方式，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，所以它将是一个请求响应过程。实际上，自从发送了一些请求后，你就做了正确的事，我能够收到回应。但为了提高效率，您的服务器也应该像布尔指示器那样做出响应。假设它已成功注册，它将响应1，如果没有，则返回0。然后您的移动应用程序将处理结果。

此外，您很可能无法获得该失败消息，因为它在try-catch之外。

android - 无法将数据发送到Web服务器的数据库

3 个答案: