这里是 4 EditText 用于输入数字密码。我希望它是这样的，如果第一个 EditText 被 1 个数字填充，那么焦点应该转到下一个 EditText 并且也应该以相反的方式工作。这样用户可以从最左边继续输入密码，也可以从最右边删除相同的方式。
有人可以建议最好的方法吗？
【问题讨论】：
标签： android android-layout android-edittext android-input-method
您无法单独使用 addTextChangedListener 来完成它。您可能必须同时设置 onKeyListener 。这是给你的代码：
//6 EditTexts are otpEt[0], otpEt[1],...otpEt[5]
private EditText[] otpEt = new EditText[6];
otpEt[0] = (EditText) findViewById(R.id.otpEt1);
otpEt[1] = (EditText) findViewById(R.id.otpEt2);
otpEt[2] = (EditText) findViewById(R.id.otpEt3);
otpEt[3] = (EditText) findViewById(R.id.otpEt4);
otpEt[4] = (EditText) findViewById(R.id.otpEt5);
otpEt[5] = (EditText) findViewById(R.id.otpEt6);
setOtpEditTextHandler();
private void setOtpEditTextHandler () { //This is the function to be called
for (int i = 0;i < 6;i++) { //Its designed for 6 digit OTP
final int iVal = i;
otpEt[iVal].addTextChangedListener(new TextWatcher() {
@Override
public void beforeTextChanged(CharSequence s, int start, int count, int after) {
}
@Override
public void onTextChanged(CharSequence s, int start, int before, int count) {
}
@Override
public void afterTextChanged(Editable s) {
if(iVal == 5 && !otpEt[iVal].getText().toString().isEmpty()) {
otpEt[iVal].clearFocus(); //Clears focus when you have entered the last digit of the OTP.
} else if (!otpEt[iVal].getText().toString().isEmpty()) {
otpEt[iVal+1].requestFocus(); //focuses on the next edittext after a digit is entered.
}
}
});
otpEt[iVal].setOnKeyListener(new View.OnKeyListener() {
@Override
public boolean onKey(View v, int keyCode, KeyEvent event) {
if (event.getAction() != KeyEvent.ACTION_DOWN) {
return false; //Dont get confused by this, it is because onKeyListener is called twice and this condition is to avoid it.
}
if(keyCode == KeyEvent.KEYCODE_DEL &&
otpEt[iVal].getText().toString().isEmpty() && iVal != 0) {
//this condition is to handel the delete input by users.
otpEt[iVal-1].setText("");//Deletes the digit of OTP
otpEt[iVal-1].requestFocus();//and sets the focus on previous digit
}
return false;
}
});
}
}
如果您觉得这段代码很难，只需将其粘贴到您的项目中并尝试重新排列即可。您将能够轻松获得它
【讨论】：
otpEt.length 和 (otpEt .lengt - 1)，它将适用于任何动态的 edtitexts。
如果您熟悉RxJava，那么这可能是满足您需求的最简单方法。
这是Kotlin 代码示例
RxTextView.textChanges(edtOtp1).filter { it.length == 1 }.subscribe { edtOtp2.requestFocus() }
RxTextView.textChanges(edtOtp2).filter { it.length == 1 }.subscribe { edtOtp3.requestFocus() }
RxTextView.textChanges(edtOtp3).filter { it.length == 1 }.subscribe { edtOtp4.requestFocus() }
RxTextView.textChanges(edtOtp4).filter { it.length == 1 }.subscribe { context.hideKeyboard(view) }
你也可以用同样的方式写反向。虽然长度为零，但对上一个 Edittext 的关注。
【讨论】：
您可以使用库Android PinView / OtpView
或者您可以使用addTextChangedListener 添加一个TextWatcher，每当此EditTextView 的文本发生更改时都会调用它，然后您可以在下一个EditText 上调用View.requestFocus() 来聚焦它
【讨论】：
你可以这样做
<LinearLayout
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:orientation="horizontal"
>
<EditText
android:id="@+id/otpET1"
android:layout_width="30dp"
android:layout_height="30dp"
android:inputType="number"
android:maxLength="1"
android:gravity="center"
android:textSize="20sp"/>
<EditText
android:id="@+id/otpET2"
android:layout_width="30dp"
android:layout_height="30dp"
android:inputType="number"
android:maxLength="1"
android:gravity="center"
android:textSize="20sp"/>
<EditText
android:id="@+id/otpET3"
android:layout_width="30dp"
android:layout_height="30dp"
android:inputType="number"
android:maxLength="1"
android:gravity="center"
android:textSize="20sp"/>
<EditText
android:id="@+id/otpET4"
android:layout_width="30dp"
android:layout_height="30dp"
android:inputType="number"
android:maxLength="1"
android:gravity="center"
android:textSize="20sp"/>
<EditText
android:id="@+id/otpET5"
android:layout_width="30dp"
android:layout_height="30dp"
android:inputType="number"
android:maxLength="1"
android:gravity="center"
android:textSize="20sp"/>
<EditText
android:id="@+id/otpET6"
android:layout_width="30dp"
android:layout_height="30dp"
android:inputType="number"
android:gravity="center"
android:maxLength="1"
android:textSize="20sp"/>
</LinearLayout>
活动中
EditText[] otpETs = new EditText[6];
otpETs[0] = findViewById(R.id.otpET1);
otpETs[1] = findViewById(R.id.otpET2);
otpETs[2] = findViewById(R.id.otpET3);
otpETs[3] = findViewById(R.id.otpET4);
otpETs[4] = findViewById(R.id.otpET5);
otpETs[5] = findViewById(R.id.otpET6);
@Override
public boolean dispatchKeyEvent(KeyEvent event) {
int keyCode = event.getKeyCode();
if (keyCode == 7 || keyCode == 8 ||
keyCode == 9 || keyCode == 10 ||
keyCode == 11 || keyCode == 12 ||
keyCode == 13 || keyCode == 14 ||
keyCode == 15 || keyCode == 16 ||
keyCode == 67) {
if (event.getAction() == KeyEvent.ACTION_DOWN) {
if (keyCode == KEYCODE_DEL) {
int index = checkWhoHasFocus();
if (index != 123) {
if (Helpers.rS(otpETs[index]).equals("")) {
if (index != 0) {
otpETs[index - 1].requestFocus();
}
} else {
return super.dispatchKeyEvent(event);
}
}
} else {
int index = checkWhoHasFocus();
if (index != 123) {
if (Helpers.rS(otpETs[index]).equals("")) {
return super.dispatchKeyEvent(event);
} else {
if (index != 5) {
otpETs[index + 1].requestFocus();
}
}
}
return super.dispatchKeyEvent(event);
}
}
} else {
return super.dispatchKeyEvent(event);
}
return true;
}
private int checkWhoHasFocus() {
for (int i = 0; i < otpETs.length; i++) {
EditText tempET = otpETs[i];
if (tempET.hasFocus()) {
return i;
}
}
return 123;
}
这只是从editTexts获取字符串
public class Helpers {
public static String rS(EditText editText) {
return editText.getText().toString().trim();
}
}
最后，
String code = Helpers.rS(otpETs[0]) + Helpers.rS(otpETs[1]) +
Helpers.rS(otpETs[2]) + Helpers.rS(otpETs[3]) + Helpers.rS(otpETs[4])
+ Helpers.rS(otpETs[5]);
或者只使用一个简单的for/while 循环。
【讨论】：
在 Kotlin 中，您可以像 .. 一样使用 bellow
txtOTP_1.addTextChangedListener(object : TextWatcher {
override fun onTextChanged(s: CharSequence, start: Int, before: Int, count: Int) {
if (txtOTP_1.text.toString().length == 1) {
txtOTP_1.clearFocus()
txtOTP_2.requestFocus()
txtOTP_2.setCursorVisible(true)
}
}
override fun beforeTextChanged(s: CharSequence, start: Int, count: Int, after: Int) {
}
override fun afterTextChanged(s: Editable) {
if (txtOTP_1.text.toString().length == 0) {
txtOTP_1.requestFocus()
}
}
})
txtOTP_2.addTextChangedListener(object : TextWatcher {
override fun onTextChanged(s: CharSequence, start: Int, before: Int, count: Int) {
if (txtOTP_2.text.toString().length == 1) {
txtOTP_2.clearFocus()
txtOTP_3.requestFocus()
txtOTP_3.setCursorVisible(true)
}
}
override fun beforeTextChanged(s: CharSequence, start: Int, count: Int, after: Int) {
}
override fun afterTextChanged(s: Editable) {
if (txtOTP_2.text.toString().length == 0) {
txtOTP_2.requestFocus()
}
}
})
txtOTP_3.addTextChangedListener(object : TextWatcher {
override fun onTextChanged(s: CharSequence, start: Int, before: Int, count: Int) {
if (txtOTP_3.text.toString().length == 1) {
txtOTP_3.clearFocus()
txtOTP_4.requestFocus()
txtOTP_4.setCursorVisible(true)
}
}
override fun beforeTextChanged(s: CharSequence, start: Int, count: Int, after: Int) {
}
override fun afterTextChanged(s: Editable) {
if (txtOTP_3.text.toString().length == 0) {
txtOTP_3.requestFocus()
}
}
})
txtOTP_4.addTextChangedListener(object : TextWatcher {
override fun onTextChanged(s: CharSequence, start: Int, before: Int, count: Int) {
if (txtOTP_4.text.toString().length == 1) {
txtOTP_4.clearFocus()
txtOTP_5.requestFocus()
txtOTP_5.setCursorVisible(true)
}
}
override fun beforeTextChanged(s: CharSequence, start: Int, count: Int, after: Int) {
}
override fun afterTextChanged(s: Editable) {
if (txtOTP_4.text.toString().length == 0) {
txtOTP_4.requestFocus()
}
}
})
txtOTP_5.addTextChangedListener(object : TextWatcher {
override fun onTextChanged(s: CharSequence, start: Int, before: Int, count: Int) {
if (txtOTP_5.text.toString().length == 1) {
txtOTP_5.requestFocus()
txtOTP_5.setCursorVisible(true)
}
}
})
【讨论】：