Question

我有一个简单的消息传递应用程序，它获取editText框的文本并将其发送到服务器。我想要做的就是当文本被发送到服务器时，我希望editText框重置。

这是我的代码，但它不会重置editText框：

public void sendMessage(View v) {
        Editable messagetext;

        messagetext = message.getText();

        final SharedPreferences prefs = PreferenceManager
                .getDefaultSharedPreferences(getBaseContext());

        username = prefs.getString("username", "null");

        where = prefs.getString("chat", "null");
        message = (EditText) findViewById(R.id.inptbox);

        function = new Functions();

        response = function
                .sendMessage(username, where, messagetext.toString());

    }

如果我再添加一行代码，为了重置该框，我的应用程序就会死掉：

public void sendMessage(View v) {
    Editable messagetext;

    messagetext = message.getText();
        message.setText("");
    final SharedPreferences prefs = PreferenceManager
            .getDefaultSharedPreferences(getBaseContext());

    username = prefs.getString("username", "null");

    where = prefs.getString("chat", "null");
    message = (EditText) findViewById(R.id.inptbox);

    function = new Functions();

    response = function
            .sendMessage(username, where, messagetext.toString());

}

我得到的错误（忍受我，我不是那么好的logcat）是：

E/AndroidRuntime(7207): java.lang.IllegalStateException: Could not execute method of the activity

全局变量列表：

String username = "";
    Functions function;
    EditText message;
    String response = "";
    String where = "";
    String inboxx;

Answer 1

...
messagetext = message.getText();                 
...
message = (EditText) findViewById(R.id.inptbox);  
...

这些顺序不应该被颠倒吗？您应该在致电getText之前说明“消息”是什么。

OnClick方法杀死android应用程序

1 个答案: