Question

我不明白为什么变量AUX和resultado不返回它们的相同值...我想保留该值以将它们返回到另一个类。你能帮帮我吗？

我的课程是：

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

    @Override
    protected String doInBackground(String... urls) {

      String response = "";
      try {

            Class.forName("com.mysql.jdbc.Driver");  
            System.out.println("driver conectado");
            Connection con = DriverManager.getConnection("jdbc:mysql://10.0.2.2:3306/savetime", "root", "root");
            resultado = "Database connection success\n";
            Statement st = con.createStatement();
            ResultSet rs = st.executeQuery("select * from ingresso");
            //ResultSetMetaData rsmd = rs.getMetaData();

            while(rs.next()) {

                resultado = rs.getString(1);
            }

            AUX = resultado ;
            System.out.println("Auxiliar 1: " + AUX + " e " + resultado);
      }
      catch(Exception e) {
          e.printStackTrace();
          System.out.println("ERRO: " + e.toString());
      }
    return response;
    }

    @Override
    protected void onPostExecute(String resultado) {
        AUX = resultado ;
        System.out.println("Auxiliar 2: "+AUX + " e " + resultado);
    }
}

logcat：

06-20 05:00:47.475: I/System.out(1592): Auxiliar 1: 1234 e 1234
06-20 05:00:47.475: I/System.out(1592): Auxiliar:  e

Answer 1

因为您的doInBackground函数返回response变量值为空。所以将AUX添加到response。试试这个

@Override
    protected String doInBackground(String... urls) {

      String response = "";
      try {

            Class.forName("com.mysql.jdbc.Driver");  
            System.out.println("driver conectado");
            Connection con = DriverManager.getConnection("jdbc:mysql://10.0.2.2:3306/savetime", "root", "root");
            resultado = "Database connection success\n";
            Statement st = con.createStatement();
            ResultSet rs = st.executeQuery("select * from ingresso");
            //ResultSetMetaData rsmd = rs.getMetaData();

            while(rs.next()) {

                resultado = rs.getString(1);
            }

            AUX = resultado ;
            response = AUX;  // Add here
            System.out.println("Auxiliar 1: " + AUX + " e " + resultado);
      }
      catch(Exception e) {
          e.printStackTrace();
          System.out.println("ERRO: " + e.toString());
      }
    return response;
}

Answer 2

在你的doInBackground方法中，你正在初始化一个响应字符串，但你从来没有为它分配AUX或者resultado，这就是为什么postExecute resultado是空的（因为它有相同的名字并不是它是相同的变量）。由于您尝试返回多个String，因此需要调整asynctask以返回包含要返回的两个值的数组或映射

我怎样才能保持相同的价值？

2 个答案: