java - '等于( )' between objects of inconvertible types ' int' 和 'TextWatcher' 错误

Question

我有一个无法解决的奇怪错误。错误是:不可转换类型“int”和“TextWatcher”的对象之间的“equals()”。这是我的代码:

if( textWatcher_ans.equals(R.string.editText_7)){
    Toast.makeText(getApplication().getBaseContext(),
                   (R.string.Good),Toast.LENGTH_SHORT).show();
}

            private final TextWatcher textWatcher_ans = new TextWatcher(){
            public void beforeTextChanged(CharSequence s, int start, int count, int after) {

            }
            public void onTextChanged(CharSequence s, int start, int before, int count) {
                textView_name.setVisibility(View.VISIBLE);
            }
            public void afterTextChanged(Editable s) {
                if (s.length() == 0) {
                    textView_name.setVisibility(View.GONE);
                } else {
                    textView_name.setText("You have entered : " + editText_surname.getText());
                }
            }

我很困惑，我不知道该怎么办。有人可以帮助我吗？

Answer 1

如果您的目标是在 editText_surname 中的文本与 R.string.editText_7 匹配时显示 R.string.Good 消息，请尝试像这样的东西:

首先将 TextWatcher 附加到 editText_surname:

editText_surname.addTextChangedListener(textWatcher_ans);

在您的afterTextChangedMethod中:

public void afterTextChanged(Editable s) {
    if (s.length() == 0) {
        textView_name.setVisibility(View.GONE);
     } else {
        textView_name.setText("You have entered : " + s.toString());
     }

    if (s.toString().equals(context.getString(R.string.editText_7)) {
        Toast.makeText(getApplication().getBaseContext(),
               (R.string.Good),Toast.LENGTH_SHORT).show();
    }
}