Javascript – Regex for Mobile Number with or without Country Code

javascriptregexvalidation

I need to validate a text field in my registration form for which I want a Regex.

The textbox will accept a mobile number like 9999999999(10 digit – starting with 7 or 8 or 9).
The user may alternatively write +919999999999, where +91 is country code (+91 for India, but can accept any other like +110 or +52), or he may also write 09999999999 (first 0 for own country)

Here, the user have 3 choices,

adding the mobile number with country code
simply without country code or 0 prefixed
With a zero prefixed in mobile number.

Though not required, my page is in asp.net, and I am using built-in regular expression validator.

Best Answer

From what I can see, this should work. The prefix is optional and is stored into the first match group, with the main number going into the second group.

^([0|\+[0-9]{1,5})?([7-9][0-9]{9})$

But if you can give us some test cases for each, it'd help us in giving you a working regex for what you want.

Props to SchlaWiener in the comments for the correct limit on the country code length.

Modern jQuery

Use .prop():

$('.myCheckbox').prop('checked', true);
$('.myCheckbox').prop('checked', false);

DOM API

If you're working with just one element, you can always just access the underlying HTMLInputElement and modify its .checked property:

$('.myCheckbox')[0].checked = true;
$('.myCheckbox')[0].checked = false;

The benefit to using the .prop() and .attr() methods instead of this is that they will operate on all matched elements.

jQuery 1.5.x and below

The .prop() method is not available, so you need to use .attr().

$('.myCheckbox').attr('checked', true);
$('.myCheckbox').attr('checked', false);

Note that this is the approach used by jQuery's unit tests prior to version 1.6 and is preferable to using $('.myCheckbox').removeAttr('checked'); since the latter will, if the box was initially checked, change the behaviour of a call to .reset() on any form that contains it – a subtle but probably unwelcome behaviour change.

For more context, some incomplete discussion of the changes to the handling of the checked attribute/property in the transition from 1.5.x to 1.6 can be found in the version 1.6 release notes and the Attributes vs. Properties section of the .prop() documentation.

Javascript – How to print a number with commas as thousands separators in JavaScript

I used the idea from Kerry's answer, but simplified it since I was just looking for something simple for my specific purpose. Here is what I did:

function numberWithCommas(x) {
    return x.toString().replace(/\B(?=(\d{3})+(?!\d))/g, ",");
}

function numberWithCommas(x) {
    return x.toString().replace(/\B(?<!\.\d*)(?=(\d{3})+(?!\d))/g, ",");
}

function test(x, expect) {
    const result = numberWithCommas(x);
    const pass = result === expect;
    console.log(`${pass ? "✓" : "ERROR ====>"} ${x} => ${result}`);
    return pass;
}

let failures = 0;
failures += !test(0,        "0");
failures += !test(100,      "100");
failures += !test(1000,     "1,000");
failures += !test(10000,    "10,000");
failures += !test(100000,   "100,000");
failures += !test(1000000,  "1,000,000");
failures += !test(10000000, "10,000,000");
if (failures) {
    console.log(`${failures} test(s) failed`);
} else {
    console.log("All tests passed");
}

.as-console-wrapper {
    max-height: 100% !important;
}

The regex uses 2 lookahead assertions:

a positive one to look for any point in the string that has a multiple of 3 digits in a row after it,
a negative assertion to make sure that point only has exactly a multiple of 3 digits. The replacement expression puts a comma there.

For example, if you pass it 123456789.01, the positive assertion will match every spot to the left of the 7 (since 789 is a multiple of 3 digits, 678 is a multiple of 3 digits, 567, etc.). The negative assertion checks that the multiple of 3 digits does not have any digits after it. 789 has a period after it so it is exactly a multiple of 3 digits, so a comma goes there. 678 is a multiple of 3 digits but it has a 9 after it, so those 3 digits are part of a group of 4, and a comma does not go there. Similarly for 567. 456789 is 6 digits, which is a multiple of 3, so a comma goes before that. 345678 is a multiple of 3, but it has a 9 after it, so no comma goes there. And so on. The \B keeps the regex from putting a comma at the beginning of the string.

@neu-rah mentioned that this function adds commas in undesirable places if there are more than 3 digits after the decimal point. If this is a problem, you can use this function:

function numberWithCommas(x) {
    var parts = x.toString().split(".");
    parts[0] = parts[0].replace(/\B(?=(\d{3})+(?!\d))/g, ",");
    return parts.join(".");
}

function numberWithCommas(x) {
    var parts = x.toString().split(".");
    parts[0] = parts[0].replace(/\B(?=(\d{3})+(?!\d))/g, ",");
    return parts.join(".");
}

function test(x, expect) {
    const result = numberWithCommas(x);
    const pass = result === expect;
    console.log(`${pass ? "✓" : "ERROR ====>"} ${x} => ${result}`);
    return pass;
}

let failures = 0;
failures += !test(0              , "0");
failures += !test(0.123456       , "0.123456");
failures += !test(100            , "100");
failures += !test(100.123456     , "100.123456");
failures += !test(1000           , "1,000");
failures += !test(1000.123456    , "1,000.123456");
failures += !test(10000          , "10,000");
failures += !test(10000.123456   , "10,000.123456");
failures += !test(100000         , "100,000");
failures += !test(100000.123456  , "100,000.123456");
failures += !test(1000000        , "1,000,000");
failures += !test(1000000.123456 , "1,000,000.123456");
failures += !test(10000000       , "10,000,000");
failures += !test(10000000.123456, "10,000,000.123456");
if (failures) {
    console.log(`${failures} test(s) failed`);
} else {
    console.log("All tests passed");
}

.as-console-wrapper {
    max-height: 100% !important;
}

@t.j.crowder pointed out that now that JavaScript has lookbehind (support info), it can be solved in the regular expression itself:

function numberWithCommas(x) {
    return x.toString().replace(/\B(?<!\.\d*)(?=(\d{3})+(?!\d))/g, ",");
}

function numberWithCommas(x) {
    return x.toString().replace(/\B(?<!\.\d*)(?=(\d{3})+(?!\d))/g, ",");
}

function test(x, expect) {
    const result = numberWithCommas(x);
    const pass = result === expect;
    console.log(`${pass ? "✓" : "ERROR ====>"} ${x} => ${result}`);
    return pass;
}

let failures = 0;
failures += !test(0,               "0");
failures += !test(0.123456,        "0.123456");
failures += !test(100,             "100");
failures += !test(100.123456,      "100.123456");
failures += !test(1000,            "1,000");
failures += !test(1000.123456,     "1,000.123456");
failures += !test(10000,           "10,000");
failures += !test(10000.123456,    "10,000.123456");
failures += !test(100000,          "100,000");
failures += !test(100000.123456,   "100,000.123456");
failures += !test(1000000,         "1,000,000");
failures += !test(1000000.123456,  "1,000,000.123456");
failures += !test(10000000,        "10,000,000");
failures += !test(10000000.123456, "10,000,000.123456");
if (failures) {
    console.log(`${failures} test(s) failed`);
} else {
    console.log("All tests passed");
}

.as-console-wrapper {
    max-height: 100% !important;
}

(?<!\.\d*) is a negative lookbehind that says the match can't be preceded by a . followed by zero or more digits. The negative lookbehind is faster than the split and join solution (comparison), at least in V8.

Best Answer

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Javascript – How to print a number with commas as thousands separators in JavaScript

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