Question

我试图在一些文字后弹出键盘。即在满足某些条件后。如何在android中执行此操作。即if（eT1的字符== et2的字符） eT1已经输入了，现在已经输入了et2。

我用过： eT2.addTextChangedListener（new TextWatcher（）{

            public void onTextChanged(CharSequence s, int start,int before, int count) 
            {
                // TODO Auto-generated method stub
    String oldPass = eT1.getText().toString();      if(!oldPass.contentEquals(e2.getText().toString()))
                {
                View view = this.getCurrentFocus();

if（view！= null）{
InputMethodManager imm =（InputMethodManager）getSystemService（Context.INPUT_METHOD_SERVICE）; imm.hideSoftInputFromWindow（view.getWindowToken（），0）; }

                }
            }
            public void beforeTextChanged(CharSequence s, int start,
                            int count, int after) {
                        // TODO Auto-generated method stub
            }

            public void afterTextChanged(Editable s) {
                        // TODO Auto-generated method stub

            }

        });

Answer 1

尝试这样的事情：

final EditText et1 = (EditText) findViewById(R.id.editText1);
final EditText et2 = (EditText) findViewById(R.id.editText2);

et1.setKeyListener(new KeyListener() {

    public boolean onKeyUp(View view, Editable text, int keyCode, KeyEvent event) {
        if(view.getText().equals(et2.getText()) {
            InputMethodManager imm = (InputMethodManager)getSystemService(Context.INPUT_METHOD_SERVICE);
            imm.hideSoftInputFromWindow(view.getWindowToken(), 0);

            return true;
        }

        return false;
    }
}

一些edittext输入后自动禁用android

1 个答案: