Question

在我的应用程序中，有一个注册屏幕，它发送用户的注册详细信息并将其发送到服务器，然后服务器将其存储在数据库中。

我的问题是，当我检查数据库时，有多个数据副本。我不知道多个副本是由客户端还是服务器端创建的。

我添加了数据库的屏幕截图，您可以清楚地看到有相同数据的多个副本。每当我在我的Android应用程序中按客户端的发送按钮时，两个相同的数据都存储在数据库中。我还没有使用任何循环，它仍然在数据库中存储两个相同数据的副本。

这是我的代码

public class RegisterScreen extends Activity {

    Button reg;
    TextView tv;
    EditText fn,ln,un,pwd,mob,email;
    HttpPost httppost;
    StringBuffer buffer;
    HttpResponse response;
    HttpClient httpclient;
    List<NameValuePair> nameValuePairs;
    ProgressDialog dialog = null;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_register);
        reg=(Button)findViewById(R.id.reg);
        fn=(EditText)findViewById(R.id.fn);
        ln=(EditText)findViewById(R.id.ln);
        un=(EditText)findViewById(R.id.un);
        pwd=(EditText)findViewById(R.id.pwd);
        mob=(EditText)findViewById(R.id.mob);
        email=(EditText)findViewById(R.id.email);
        tv = (TextView)findViewById(R.id.tv);


        reg.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                dialog = ProgressDialog.show(RegisterScreen.this, "",
                        "Registering user...", true);
                new Thread(new Runnable() {
                    public void run() {
                        login();
                    }
                }).start();
            }
        });
    }


    void login(){
        try{

            httpclient=new DefaultHttpClient();
            httppost= new HttpPost("http://192.168.43.6/test/registration.php"); // make sure the url is correct.
            //add your data
            nameValuePairs = new ArrayList<NameValuePair>();
            // Always use the same variable name for posting i.e the android side variable name and php side variable name should be similar,
            nameValuePairs.add(new BasicNameValuePair("firstname",fn.getText().toString().trim()));
            nameValuePairs.add(new BasicNameValuePair("lastname",ln.getText().toString().trim()));
            nameValuePairs.add(new BasicNameValuePair("username",un.getText().toString().trim()));
            nameValuePairs.add(new BasicNameValuePair("password",pwd.getText().toString().trim()));
            nameValuePairs.add(new BasicNameValuePair("mob",mob.getText().toString().trim()));
            nameValuePairs.add(new BasicNameValuePair("email",email.getText().toString().trim()));// $Edittext_value = $_POST['Edittext_value'];

            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            //Execute HTTP Post Request
            response=httpclient.execute(httppost);

            ResponseHandler<String> responseHandler = new BasicResponseHandler();
            final String response = httpclient.execute(httppost, responseHandler);
            System.out.println("Response : " + response);
            runOnUiThread(new Runnable() {
                public void run() {
                    tv.setText("Response from PHP : " + response);
                    dialog.dismiss();
                }
            });

            if(response.equalsIgnoreCase("Registration")){
                runOnUiThread(new Runnable() {
                    public void run() {
                        Toast.makeText(RegisterScreen.this,"Registration Successful", Toast.LENGTH_SHORT).show();
                    }
                });

                startActivity(new Intent(RegisterScreen.this, LoginScreen.class));
            }else{
                showAlert();
            }

        }catch(Exception e){
            dialog.dismiss();
            System.out.println("Exception : " + e.getMessage());
        }
    }
    public void showAlert(){
        RegisterScreen.this.runOnUiThread(new Runnable() {
            public void run() {
                AlertDialog.Builder builder = new AlertDialog.Builder(RegisterScreen.this);
                builder.setTitle("Registration Error.");
                builder.setMessage("Problem in Registration. Please try again later")
                        .setCancelable(false)
                        .setPositiveButton("OK", new DialogInterface.OnClickListener() {
                            public void onClick(DialogInterface dialog, int id) {
                            }
                        });
                AlertDialog alert = builder.create();
                alert.show();
            }
        });
    }
}

这是我的服务器端代码

<?php
$hostname_localhost ="localhost";
$database_localhost ="testdb";
$username_localhost ="root";
$password_localhost ="";
$localhost = mysql_connect($hostname_localhost,$username_localhost,$password_localhost)
or
trigger_error(mysql_error(),E_USER_ERROR);

mysql_select_db($database_localhost, $localhost);

$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$username = $_POST['username'];
$password = $_POST['password'];
$mob = $_POST['mob'];
$email = $_POST['email'];

$query_search = "INSERT INTO register(Firstname, Lastname, loginname, pwd, Mobilenumber , emailid) VALUES('$firstname', '$lastname','$username','$password','$mob','$email')";
$query_exec = mysql_query($query_search) or die(mysql_error());
//$rows = mysql_num_rows($query_exec);
//echo $rows;
 echo "Registration";
?>

这是数据库 enter image description here

的屏幕截图

Answer 1

好像你的调用在Login()执行了两次？

response=httpclient.execute(httppost);

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

Answer 2

我不知道多个副本是否由客户端创建或服务器端。

此代码每次运行时都会插入一个新行

$query_search = "INSERT INTO register(Firstname, Lastname, loginname, pwd, Mobilenumber , emailid) VALUES('$firstname', '$lastname','$username','$password','$mob','$email')";
$query_exec = mysql_query($query_search) or die(mysql_error());

您需要选择现有数据，然后才能插入行

Android JSON解析器将多个数据副本发送到服务器

2 个答案: