Can this function be rewritten with a regex?
Yes you can:
^(?:BE)?\s*[0-1]?(\d[. ]*){9}$
This regex should do it!
This source (in Dutch) states what an enterprise number is for Belgium:
It has the country code:
BEfollowed by a0or1and then followed by 9 digits.
https://regex101.com/r/4SRHxi/4
Explanation:
^: the string has to start with the given regex(?:BE)?: look for a group with BE but?means it matches zero or one times -?:means find but don't capture\s*: search for a space that matches zero or unlimited times[0-1]?: check if a zero of one is present zero or one times((\d[. ]*){9}): Check if 9 digits follow the remaining string, doesn't matter with how many dots or spaces they're padded. Every iteration is captured as the 1st capturing group. This becomes important when we replace later.$: the string has to end
This will check if the input validates.
Editing it into the code version is simple:
«code».replace(/^(?:BE)?\s*[0-1]?((\d[. ]*){9})$/g, function(){
return arguments[1].replace(/\D/g, "");
});
the g or global modifier will ensure all unwanted characters will be deleted. By using a function with a replace in it to replace all non-digit characters. This functions will output our desired result.
document.querySelector("pre").textContent.split("\n").forEach(function(element){
if (element.match(/^(?:BE)?\s*[0-1]?(\d[. ]*){9}$/))
{
console.log(element.replace(/^(?:BE)?\s*[0-1]?((\d[. ]*){9})$/g, function(){
return arguments[1].replace(/\D/g, "");
}));
}
else
{
console.log(`REJECTED: ${element}`);
}
});<pre>
BE 0123.321.123
BE0123.321.123
BE0123 321 123
BE 0123 321 123
BE 01 23 32 11 23
BE 0123 32 11 23
1123.321.123
123.321.123
123321123
AAA3434343A
BE 1233 445 4545 442
</pre>Rebuilding the String into the correct user friendly way is easy now:
document.querySelector("pre").textContent.split("\n").forEach(function(element) {
if (element.match(/^(?:BE)?\s*[0-1]?((\d[. ]*){9})$/)) {
var stripped = element.replace(/^(?:BE)?\s*[0-1]?((\d[. ]*){9})$/g, function(){
return arguments[1].replace(/\D/g, "");
});
//with the modulo check from your code added back in.
if (97 - (parseInt(stripped.slice(0, 7), 10) % 97) == parseInt(stripped.slice(7, 9), 10)) {
//use a literal string
//use substring to put the dots between the sections of three numbers.
var humanReadable = `BE 0${stripped.substring(0,3)}.${stripped.substring(3,6)}.${stripped.substring(6,9)}`;
console.log(`CODE: ${stripped}`, `UI: ${humanReadable}`);
}
}
});<pre>
BE 0123.321.123
BE0123.321.123
BE0123 321 123
0123.321.123
123.321.123
123321123
844256524
</pre>Second Question Yes, this can be done however it requires you to write your own code for it.
Simple version:
document.querySelector("div.enterprisenumber > input").addEventListener("keydown", function(e) {
let value = this.value;
//prevent the input from going back to 0
if ( (value.length == 0 && (e.key == "Backspace" || e.key == "Delete"))) {
e.preventDefault();
return false;
}
}, true);
document.querySelector("div.enterprisenumber > input").addEventListener("keyup", function(e) {
//reset to a value without dots
let value = this.value.replace(/\./g, "");
//strip the leading zero
const valueWithout = value;
//calculate how much iterations we need of a groups of three digits.
const i = Math.floor(valueWithout.length / 3);
let newValue = "";
//check if backspace or delete are used to make sure the dot can be deleted.
if (valueWithout.length < 9 && !(e.key == "Backspace" || e.key == "Delete")) {
//only fire when higher than zero
if (i > 0) {
let t;
//t is the index
for (t = 0; t < i; t++) {
//slice the correct portion of the string and append a dot, unless we are at the end of the groups
newValue += valueWithout.slice(t * 3, t * 3 + 3) + (t == 2 ? "" : ".");
}
//append the remainder that is not a group of three.
newValue += valueWithout.slice((t) * 3);
} else {
//return the value as is.
newValue = value;
}
//set the new value to the input.
this.value = newValue;
}
}, true);
document.querySelector("div.enterprisenumber > input").addEventListener("blur", function(e) {
let passed = false;
if (this.value.match(/^(?:BE)?\s*[0-1]?((\d[. ]*){9})$/))
{
const value = this.value.replace(/\./g, "");
//with modulo check
if (97 - (parseInt(value.slice(0,7), 10) % 97) == value.slice(7, 9))
{
passed = true;
}
}
document.querySelector(".enterprisenumber").classList[(passed ? "remove" : "add")]("error");
});
//if focus then focus input
document.querySelector("div.enterprisenumber").addEventListener("click", function(e) {
if (e.target && e.target.nodeName != "SELECT")
{
this.querySelector("input").focus();
}
});* {
box-sizing: border-box;
font-family: tahoma;
font-size: 10pt;
}
div.enterprisenumber {
border: 1px solid #747474;
width: 300px;
padding: 0px;
display: grid;
grid-template-columns: 25px 40px auto;
border-radius: 10px;
}
div.enterprisenumber.error {
border: 1px solid #ff0000;
}
div.enterprisenumber>span {
grid-column: 1;
border: 0px;
padding: 5px;
background: linear-gradient(to right, rgba(0,0,0, 0.8) 33%, rgba(255,243,54, 0.8) 33%, rgba(255, 243, 54, 0.8) 66%, rgba(255, 15, 33, 0.8) 66%, rgba(255, 15, 33, 0.8) 100%);
color: #ffffff;
font-weight: bold;
text-shadow: 1px 1px #000000;
border-radius: 10px 10px 10px 10px;
}
div.enterprisenumber>select {
grid-column: 2;
border: 0px;
padding: 5px;
}
div.enterprisenumber>input {
grid-column: 3;
border: 0px;
padding: 5px;
border-radius: 0px 10px 10px 0px;
}Enter: 844256524
<div class="enterprisenumber">
<span>BE</span><select><option value="0">0</option><option value="1">1</option><input value="" maxlength="11" />
</div>For your example strings, you could match:
^(?:BE\s?)?[01]?(\d{3}([. ])\d{3}\2\d{3}|\d{9})$
That will match
^Start of string(?:BE\s?)?Optional BE followed by optional whitespace char[01]?Optional zero or 1(Capturing group\d{3}Match 3 digits([. ])Capture in group either a space or digit to use as backreference\d{3}\2\d{3}Match 3 digits, dot or space (\2 is the backreference) and 3 digits|Or\d{9}Match 9 digits
)Close capturing group$End of string
Regex demo
And then in the replacement use the first capturing group and replace the space or the dot with an empty string.
let pattern = /^(?:BE\s?)?[01]?(\d{3}([. ])\d{3}\2\d{3}|\d{9})$/;
let strings = [
"BE 0123.321.123",
"BE0123.321.123",
"BE0123 321 123",
"0123.321.123",
"123.321.123",
"123321123",
];
strings = strings.map(x => x.replace(pattern, function(m, g) {
let enterpriseNumber = g.replace(/[. ]/g, "");
return `BE 0${enterpriseNumber.substring(0, 3)}.${enterpriseNumber.substring(3, 6)}.${enterpriseNumber.substring(6, 9)}`
}));
console.log(strings);Here is an implementation of a BE ____.___.___ style of input. The pattern will be maintained, so the input will be guaranteed to have the "BE" prefix, the space, and the two dots. The validation can then concentrate on completeness and the modulo test.
Note that the input requires the first group to have 4 digits, where the first digit must be a 0 or a 1.
const ent = document.getElementById("ent");
const out = document.getElementById("isvalid");
function format() {
const re = /^\D*[2-9]+|\D+/g;
const [i, j] = [this.selectionStart, this.selectionEnd].map(i => {
i = this.value.slice(0, i).replace(re, "").length;
return i + 3 + (i >= 4 + format.backspace) + (i >= 7 + format.backspace);
});
this.value = "BE " + this.value.replace(re, "").padEnd(10, "_")
.replace(/(....)(...)(...).*/, "$1.$2.$3");
this.setSelectionRange(i, j);
format.backspace = false;
out.textContent = validate(this.value) ? "is valid" : "is invalid";
}
function validate(num) {
return /^BE [01](\d{3}\.){2}\d{3}$/.test(num)
&& 97 - num.replace(/\D/g, "").slice(0, 8) % 97 === +num.slice(-2);
}
ent.addEventListener("input", format);
ent.addEventListener("keydown", (e) => format.backspace = e.key == "Backspace");Belgian enterprise number: <input id="ent" value="BE ____.___.___">
<span id="isvalid"></span>