正则表达式替换前 4 位数字后的数字

Question

I am trying to write a Regular Expression that replaces all the digits of a number with *'s after first 4 digits.

For example

var number = 123456789  

it should be replaced with 1234*****

Answer 1

In Javascript:

var maskedNumber = String(number).substr(0,4) + Array(String(number).length - 3).join('*');

In PHP:

$maskedNumber = str_pad(substr($number, 0, 4), strlen($number), "*");

Answer 2

In PHP:

function replaceDigit_callback($matches) {
    return $matches[1] . str_repeat('*', strlen($matches[2]));
}
$text = '1234567890';
echo $text, "\n";
$text = preg_replace_callback('#(\d{4})(\d+)#', 'replaceDigit_callback', $text);
echo $text, "\n";

Output:

1234567890
1234******

In JS:

var number = 1234567890;
var output = number.toString().replace(/(\d{4})(\d*)/, function (str, p1, p2) { return p1 + p2.replace(/./g, '*') });

Answer 3

You Can set your PHP coding..I provide you logic....

using System;
using System.Text.RegularExpressions;

class RegexSubstitution
{
   public static void Main()
   {
      string testString1 = "1, 2, 3, 4, 5, 6, 7, 8";
      Regex testRegex1 = new Regex( @"\d" );

      Console.WriteLine( "Original string: " + testString1 );
      Console.WriteLine( "Replace first 3 digits by \"digit\": " + testRegex1.Replace( testString1, "digit", 3 ) );
   }
}

Answer 4

Without regex:

var number = '123456789';
var output = '';

output = number.substr(0,4);
for ( var i = 0; i < number.length - 4; ++i )
{
    output += '*';
}

Input number has to be string !

Conversion

var number = 1234567890;

to string looks just like:

var number = 1234567890 + '';

or

var number = parseString(1234567890);