<EditText
android:id="@+id/name"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:imeOptions="actionNext"
android:inputType="textNoSuggestions|textCapCharacters"
android:maxLength="3"
android:hint="ABC"
/>
当EditText
中的所有3个符号都填满时,还有一个name
应该聚焦。
name.addTextChangedListener(object: TextWatcher {
override fun afterTextChanged(s: Editable?) {
// Move to surname when all three symbols are entered.
if (name.text.toString().length == 3) {
surname.requestFocus()
}
}
override fun beforeTextChanged(s: CharSequence?, start: Int, count: Int, after: Int) { }
override fun onTextChanged(s: CharSequence?, start: Int, before: Int, count: Int) { }
})
当我们输入1、2和3个符号时,它可以正常工作。但是,当我们在name
中选择最后一个位置时,在键盘上按一个新字母,则什么也没有发生。我试图抓住一个按键
name.setOnKeyListener { v, keyCode, event ->
if (name.text.getText().length == 3) {
surname.requestFocus()
}
false
}
但不会发生任何事件。如何移动焦点?
答案 0 :(得分:1)
将maxLength
更改为4
android:maxLength="4"
在afterTextChanged
中:
override fun afterTextChanged(s: Editable?) {
// Move to surname when all three symbols are entered.
if (name.isFocused() && name.text.toString().length > 3) {
surname.requestFocus();
name.setText(s.toString().substring(0,3));
}
}
答案 1 :(得分:0)
在beforeTextChanged()中添加一个检查,以确保char序列的长度已经是3个字符,如果是,则要求关注姓氏
答案 2 :(得分:0)
根据@nupadhyaya的回答,我是这样的。
1)删除了android:maxLength="3"
(或可以设置android:maxLength="4"
)。
2)为新符号添加了一个事件:
name.addTextChangedListener(object: TextWatcher {
override fun afterTextChanged(s: Editable?) {
val text = name.text.toString()
if (text.length >= 3) {
surname.requestFocus()
if (text.length > 3) {
// Remove symbols after third.
name.setText(text.substring(0, 3))
}
}
override fun beforeTextChanged(s: CharSequence?, start: Int, count: Int, after: Int) { }
override fun onTextChanged(s: CharSequence?, start: Int, before: Int, count: Int) { }
})
}
3)为姓氏中的Backspace
键添加了一个事件:
surname.setOnKeyListener { _, keyCode, _ ->
if (keyCode == KeyEvent.KEYCODE_DEL && surname.text.toString().isEmpty()) {
// Move to the last symbol of name.
name.requestFocus()
name.setSelection(name.text.toString().length)
}
false
}