如何从输入类型编号中读取特殊字符

Question

In my form I have a few number input fields. Normally there is only one digit allowed. After that one digit has been put in, onKeyUp the next input field is selected. However for the cases where more than one digit are required I want to use '*' as an indicator. So when the users types *23 it won't skip.

I have come that far, however when I get the value of the input field I receive an empty string.

I want to remove the asterisk from the input. When I try to alter the input of the field in the keydown event, where multi is set to true, the input field value is not altered.

Altering the input type does fix the problem with the value being "" , but then the step isn't functional any more.

I feel like I'm making a giant pile of code for something I feel should be fairly simple.

<input type="number" step="1" min="0" value="0">

var multi = 0;
// Keys to be ignored by the keyup event of .singleDigit.
var ignoreKeys = [ [8,46], [65,93], [107,222] ];
// Don't accept tab button, will skip an extra input field.
// Don't skip to next input, if input starts with *
// DOn't accept function buttons
$(".singleDigit").on('keyup', function(e){
  if(multi == 0
    && $.isNumeric($(this).val()) 
    && (e.which < ignoreKeys[0][0] || e.which > ignoreKeys[0][1])
    && (e.which < ignoreKeys[1][0] || e.which > ignoreKeys[1][1])
    && e.which < ignoreKeys[2][0]) {
    var inputs = $(this).closest('form').find('.input');
    inputs.eq( inputs.index(this) + 1).focus();
  }
});
// on focus out reset multi boolean.
$(".singleDigit").on('focusout', function() {
  multi = 0;
});
// check for key combination of *
var lastKey = null;
$(".singleDigit").on('keydown', function(e){
  if(lastKey != null) {
    if(lastKey == 16 && e.which == 56) {
      multi = 1;
    }
  }
  if(e.which != null) {
    lastKey = e.which;
  }
});

Answer 1

I found a way around it. Since I have the event, I can just prevent the default action.

var lastKey = null;
$(".singleDigit").on('keydown', function(e){
  if(lastKey != null) {
    if(lastKey == 16 && e.which == 56) {
      multi = 1;
      e.preventDefault();
    } 
  }
  if(e.which != null) {
    lastKey = e.which;
  }
});