Question

我正在使用Android中的php mysql进行登录和注册活动。我试过测试注册部分，但它没有工作。我的php文件已经过测试，它们正常运行。我怀疑我的问题可能是我提供的网址，但它可以在浏览器中使用。在注册表单中输入详细信息后，单击“提交”按钮不会执行任何操作。

这是我的代码： Login.java

public class Login extends AppCompatActivity {

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_login);
    Toolbar toolbar = (Toolbar) findViewById(R.id.toolbar);
    setSupportActionBar(toolbar);


}


public void userLogin(View view)
{

}

public void userReg(View view)
{
    startActivity(new Intent(this, Registration.class));
}

}

Registration.java

public class Registration extends Activity {

EditText ETname, ETusername, ETpassword;
String name, username, userpass;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_registration);


    ETname = (EditText)findViewById(R.id.newUsername);
    ETusername = (EditText)findViewById(R.id.newUserid);
    ETpassword = (EditText)findViewById(R.id.newUserpassword);


}

public  void reg(View view)
{
    name=ETname.getText().toString();
    username=ETusername.getText().toString();
    userpass=ETpassword.getText().toString();
    String method= "register";
    BackgroundTask backgroundTask = new BackgroundTask(this);
    backgroundTask.execute(method,name,username,userpass);
    finish();

}

}

BackgroundTask.java

public class BackgroundTask extends AsyncTask<String, Void, String> {

Context ctx;
BackgroundTask(Context ctx)
{
    this.ctx=ctx;

}

@Override
protected String doInBackground(String... params)
{
    String reg_url="http://127.0.0.1/webapp/register.php";
    String login_url="http://127.0.0.1/webapp/login.php";
    String method= params[0];
    if(method.equals("register"))
    {
        String name=params[1];
        String user_name=params[2];
        String user_pass=params[3];

        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","UTF-8")+"="+URLEncoder.encode(name,"UTF-8")+"&"+
                    URLEncoder.encode("user_name","UTF-8")+"="+URLEncoder.encode(user_name,"UTF-8")+"&"+
                    URLEncoder.encode("user_pass","UTF-8")+"="+URLEncoder.encode(user_pass,"UTF-8");
            bufferedWriter.write(data);
            bufferedWriter.flush();
            bufferedWriter.close();
            os.close();
            InputStream is= httpURLConnection.getInputStream();
            is.close();
            return "Registration Success...";


        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
    return null;
}

@Override
protected void onPreExecute() {
    super.onPreExecute();
}

@Override
protected void onProgressUpdate(Void... values) {
    super.onProgressUpdate(values);
}

@Override
protected void onPostExecute(String result) {
    Toast.makeText(ctx,result,Toast.LENGTH_LONG).show();
}

}

Answer 1

您可能不需要将KEY转换为encode，只需将值encoded转换为String data = URLEncoder.encode("user","UTF-8")+"="+URLEncoder.encode(name,"UTF-8")+"&"+ URLEncoder.encode("user_name","UTF-8")+"="+URLEncoder.encode(user_name,"UTF-8")+"&"+ URLEncoder.encode("user_pass","UTF-8")+"="+URLEncoder.encode(user_pass,"UTF-8");所以，

更改此

String data = "user"="+URLEncoder.encode(name,"UTF-8")+"&"+
                        "user_name"+"="+URLEncoder.encode(user_name,"UTF-8")+"&"+"user_pass"+"="+URLEncoder.encode(user_pass,"UTF-8");

要

<form action="YourPageName.aspx" method="post">
    <input name="TextBox1" type="text" value="" id="TextBox1" />
    <input name="TextBox2" type="password" id="TextBox2" />
    <input type="submit" name="Button1" value="Button" id="Button1" />
</form>

看看这是否适合您。

Answer 2

你说你点击了提交按钮，也许你没有正确调用onClick。您是否尝试通过findViewById（R.id.button）获取活动按钮的引用？同时尝试打印网址以确保格式正确。

Android登录注册

2 个答案: