Java &引用；getText"；语法不起作用，它'；显示错误

“getText”语法不起作用，显示错误

    @Override
    protected String doInBackground(String... params)
    {
        String usernam = username.getText().toString();
        String passwordd = password.getText().toString();
        if(usernam.trim().equals("")|| passwordd.trim().equals(""))
        {
            z = "Please enter Username and Password";
        }
        else
        {
            try
            {
                con = connectionclass(un, pass, db, ip);
                if (con == null)
                {
                    z = "Check Tour Internet Access!";
                }
                else
                {
                    String query = "select * from login where user_name= '" + usernam.toString() + "' and pass_word = ' " + passwordd.toString();
                    Statement stmt = con.createStatement();
                    ResultSet rs = stmt.executeQuery(query);
                    if(rs.next())
                    {
                        z = "Login successful";
                        isSuccess = true;
                        con.close();
                    }
                    else
                    {
                        z = "Invalid Credentials";
                        isSuccess = false;
                    }
                }
            }
            catch (Exception ex)
            {
                isSuccess = false;
                z = ex.getMessage();
            }
        }
        return z;
    }

我无法得到答案，预期的答案是，应该能够检索用户名、密码并将其与我们输入的用户名密码匹配，但现在我无法同时匹配这两个

private class MyAsyncTask extends AsyncTask<String, Void, String> {

 private String username="";
 private String password="";

 public MyAsyncTask(String user, String pass)
 {
    this.username = user;
    this.password = pass;
 }

 protected void onPreExecute() {

 }

 protected Bitmap doInBackground(String... strings) {
     // Some long-running task like downloading an image.
     // No UI related work here.. or else it will crash

 }



 protected void onPostExecute(Bitmap result) {
     // This method is executed in the UIThread
     // with access to the result of the long running task

 }
}

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在活动中调用AsyncTask本身时传递EditText的值

什么不起作用？你期待什么？相反会发生什么？请解释。顺便说一句，在创建SQL查询时不要使用连接，否则您将引入。改为使用PreparedStatement.String usernam=username.getText（）.toString（）；字符串passwordd=password.getText（）.toString（）；这两行显示了错误您得到了什么错误？无法解析方法。getText（）error@Paul P Jobyyou应该将该方法与您在活动中使用的textView名称一起使用。您应该使用findViewById（R.id.nameoftextview）在主活动中声明用户名和密码的textView方法如果您认为问题在于@Tharminian and SivanesAnsorry login=（按钮）findViewById（R.id.btnlogin），请批准答案；用户名=（EditText）findViewById（R.id.editText2）；密码=（EditText）findViewById（R.id.EditText）；progressBar=（progressBar）findViewById（R.id.progressBar）；你是说这部分吗

// Its inside the Activity Class 
MyAsyncTask asd = new MyAsyncTask(username.getText().toString(),password.getText().toString());

asd.execute("");