在Android中,如何针对EditText禁用“登录”按钮？

如果EditText为空,则必须禁用“登录按钮”.如果EditText有一些文本,则必须启用登录按钮.那么你可以在Instagram登录上看到这个方法.

两个字段都是空的,登录按钮是禁用的.

密码字段为空,因此仍然禁用Sign inButton.

这里Username和Password字段都不为空,所以Sign inButton是启用的.

如何实现这些步骤？
这是我的代码,它不起作用..

EditText et1,et2;
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_login_check);
    et1 = (EditText) findViewById(R.id.editText1);
    et2 = (EditText) findViewById(R.id.editText2);
    Button b = (Button) findViewById(R.id.button1);

    String s1 = et1.getText().toString();
    String s2 = et2.getText().toString();

    if(s1.equals("")|| s2.equals("")){
        b.setEnabled(false);
    } else {
        b.setEnabled(true);
    }
}

解决方法:

你还在寻找什么：

private EditText et1,et2;
//  create a textWatcher member
private TextWatcher mTextWatcher = new TextWatcher() {
    @Override
    public void beforeTextChanged(CharSequence charSequence, int i, int i2, int i3) {
    }

    @Override
    public void onTextChanged(CharSequence charSequence, int i, int i2, int i3) {
    }

    @Override
    public void afterTextChanged(Editable editable) {
        // check Fields For Empty Values
        checkFieldsForEmptyValues();
    }
};

void checkFieldsForEmptyValues(){
    Button b = (Button) findViewById(R.id.button1);

    String s1 = et1.getText().toString();
    String s2 = et2.getText().toString();

    if(s1.equals("")|| s2.equals("")){
        b.setEnabled(false);
    } else {
        b.setEnabled(true);
    }
}

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_login_check);
    et1 = (EditText) findViewById(R.id.editText1);
    et2 = (EditText) findViewById(R.id.editText2);


    // set listeners
    et1.addTextChangedListener(mTextWatcher);
    et2.addTextChangedListener(mTextWatcher);

    // run once to disable if empty
    checkFieldsForEmptyValues(); 
}

标签：android,android-edittext,button,android-ui,android-button
来源： https://codeday.me/bug/20191004/1852251.html