Regex Replace variable number of ****** with predefined digits

Question

var panmaskednumber = "543034******0243"; Console.WriteLine(panmaskednumber.Count(x => x == '*'));
var pattern = "\\*";
var replace = "123456789";
Regex reg = new Regex(pattern);
var newnumber = reg.Replace(panmaskednumber, replace,panmaskednumber.Count(x => x == '*'));
Console.WriteLine(newnumber);

I'm trying to Replace * in var panmaskednumber(coming from DB with symmetric key).

Not liking to use the Contains approach in which I'm specifying number of * 6 / 7 with multiple If-elseif. Since those can vary between 6,7-9.

With my above approach it replaces for each char of -> * with var replace. Any Linq approach if there is highly appreciated. Result something: 5430341234567890243

What sould be the output if *'s count is greater that replace's length? — aloisdg
– aloisdg, Commented Apr 12, 2017 at 13:03
You should be more specific about the input and desired output. "Result something" did not make clear the dependency between the number of * and the digits you expect as a replacement. — Filburt
– Filburt, Commented Apr 12, 2017 at 13:08

Wiktor Stribiżew · Accepted Answer · 2017-04-12 13:13:05Z

3

You need to use a \*+ pattern that will match 1 or more asterisk symbols:

var panmaskednumber = "543034******0243";
var replace = "123456789";
var res = Regex.Replace(panmaskednumber, @"\*+", replace);
// res => 5430341234567890243

See the C# demo.

If the number of asterisks to be replaced depends on the replace length, you may pass the match value to the match evaluator and perform necessary manipulatons there:

var panmaskednumber = "543034*****0243";
var replace = "123";

var res = Regex.Replace(panmaskednumber, @"\*+", m => 
      m.Value.Length <= replace.Length ? 
        replace.Substring(0, m.Value.Length) : 
        $"{replace}{m.Value.Substring(replace.Length)}"
    );
Console.Write(res);
// "543034***0243" / "123456789" -> 543034 123 0243
// "543034*****0243" / "123" -> 543034 123** 0243

See antother C# demo

edited Apr 12, 2017 at 13:13

answered Apr 12, 2017 at 12:54

Wiktor Stribiżew

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8 Comments

Filburt Over a year ago

The expected out would be 5430341234560243 - the OP wants to replace the exact number of asterisk symbols with the matching number of digits.

Rawling Over a year ago

@Filburt That's not what he says

Wiktor Stribiżew Over a year ago

Well, if the number of asterisks to be matches depends on the replace length, it still can be managed. Let me update.

Filburt Over a year ago

@Rawling OP has some difficulties explaing but states "Since those can vary between 6,7-9." and "With my above approach it replaces for each char of -> * with var replace."

Wiktor Stribiżew Over a year ago

@aloisdg: Yes, $"{replace}{m.Value.Substring(replace.Length)}" will work.

|

aloisdg · Accepted Answer · 2017-04-12 13:21:32Z

You can use a Regex, but you can achieve without. Lets go for a simpler solution. Try it online.

var panmaskednumber = "543034******0243";
var count = panmaskednumber.Count(x => x == '*');
var start = panmaskednumber.IndexOf('*');   
var replace = "123456789";

// output 5430341234567890243 (543034 123456789 0243)
Console.WriteLine(panmaskednumber.Remove(start) // get head
                + replace // add replace
                + panmaskednumber.Substring(start + count)); // add tail

 // output 5430341234560243 (543034 123456 0243) // get head 
 Console.WriteLine(panmaskednumber.Remove(start)
                + replace.Remove(count) // add replace with count respect
                + panmaskednumber.Substring(start + count)); // add tail

 replace = "123";

 // output 543034123***0243 (543034 123*** 0243) // get head 
 Console.WriteLine(panmaskednumber.Remove(start)
                + replace // add replace
                + new string('*', count - replace.Length) // fill with missing *
                + panmaskednumber.Substring(start + count)); // add tail

"I suppose it is tempting, if the only tool you have is a hammer, to treat everything as if it were a nail."

Law of instrument

Dont use Regex, if you don't have too. For this problem, C#.NET is enough. :)

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