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[英]json schema not working for type = number and pattern = 10 digit regular pattern
[英]how to make User Input of 10 digit Mobile number into 3 - 3 - 4 pattern
如何使用戶將10位手機號碼輸入為3-3-4格式? 9840098480到(984)-809-8480在android中的示例?
只需使用PhoneNumberFormattingTextWatcher ,只需調用:
editText.addTextChangedListener(new PhoneNumberFormattingTextWatcher());
加成
需要明確的是,PhoneNumberFormattingTextWatcher的主干是PhoneNumberUtils類。 區別在於TextWatcher維護EditText,而每次更改其內容時必須調用PhoneNumberUtils.formatNumber()
。
要么
您可以使用此:
<EditText
android:id="@+id/editText1"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:inputType="phone" />
然后嘗試以下代碼:
final EditText text = (EditText) findViewById(com.and.R.id.editText1);
text.addTextChangedListener(new TextWatcher() {
@Override
public void onTextChanged(CharSequence s, int start, int before, int count) {
boolean flag = true;
String eachBlock[] = text.getText().toString().split("-");
for (int i = 0; i < eachBlock.length; i++) {
if (eachBlock[i].length() > 4) {
flag = false;
}
}
if (flag) {
text.setOnKeyListener(new OnKeyListener() {
@Override
public boolean onKey(View v, int keyCode, KeyEvent event) {
if (keyCode == KeyEvent.KEYCODE_DEL)
keyDel = 1;
return false;
}
});
if (keyDel == 0) {
if (((text.getText().length() + 1) % 5) == 0) {
if (text.getText().toString().split("-").length <= 3) {
text.setText(text.getText() + "-");
text.setSelection(text.getText().length());
}
}
a = text.getText().toString();
} else {
a = text.getText().toString();
keyDel = 0;
}
} else {
text.setText(a);
}
}
@Override
public void beforeTextChanged(CharSequence s, int start, int count,
int after) {
// TODO Auto-generated method stub
}
@Override
public void afterTextChanged(Editable s) {
}
});
使用Himanshu Agarwal的方法或執行以下操作:
String number = "1234567899";
System.out.println("(" + number.substring(0, 3) + ")-"
+ number.substring(3, 6) + "-" + number.substring(6));
輸出繼電器:
(123)-456-7899
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