Question

这里我初始化在btnlog onclick

之后获取用户输入的值

protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

    usernameField = (EditText) findViewById(R.id.etUname);
    passwordField = (EditText) findViewById(R.id.etUpass);

    btnlog = (Button) findViewById(R.id.loginbtn);
    btnview = (Button) findViewById(R.id.viewbtn);
    reglink = (TextView) findViewById(R.id.reglink);

    btnlog.setOnClickListener (new View.OnClickListener(){
        @Override
        public void onClick(View v){

            String u = usernameField.getText().toString();
            String p = passwordField.getText().toString();

            new loginPost().execute();
        }
    });

现在我想从公共类接收用户输入并对其进行编码，变量u和p基本上没有从公共类中获取任何值

@Override
        protected String doInBackground(String... params) {
            try{

            String str = "username=" + u + "&password=" + p;
            String data = URLEncoder.encode("str", "UTF-8") + "=" +
                    URLEncoder.encode(str, "UTF-8");
            strUrl ="http://10.0.2.2/android/login.php?"+data+"";
            Log.d("STR", data);


            URL url = new URL (strUrl);
            HttpURLConnection con = (HttpURLConnection) url.openConnection();
            con.setRequestMethod("POST");
            con.connect();

Answer 1

在onCreate（）之外定义u和p并将其声明为public。现在您可以通过

从其他类访问这些变量

    this.p=MainActivity.p;
    this.u=MainActivity.u;

Answer 2

我的java不是很流畅但看起来你试图将变量传递给asynctask，如果是这样的话那么看看这里， How to pass variables in and out of AsyncTasks? 它显示了如何在.execute（）调用

上传递变量

Answer 3

您可以使用.execute()方法传递它，因为这就是设计。

btnlog.setOnClickListener (new View.OnClickListener(){
        @Override
        public void onClick(View v){
            String u = usernameField.getText().toString();
            String p = passwordField.getText().toString();
            new loginPost().execute(u,p);
        }
});

然后在doInBackground()

中

@Override
protected String doInBackground(String... params) {
    try{

    String str = "username=" + params[0] + "&password=" + params[1];
    String data = URLEncoder.encode("str", "UTF-8") + "=" +
            URLEncoder.encode(str, "UTF-8");
    strUrl ="http://10.0.2.2/android/login.php?"+data+"";
    Log.d("STR", data);


    URL url = new URL (strUrl);
    HttpURLConnection con = (HttpURLConnection) url.openConnection();
    con.setRequestMethod("POST");
    con.connect();

android studio：如何从另一个类

3 个答案: