How to remove leading zeros using C#

Question

This is the code you need:

string strInput = "0001234";
strInput = strInput.TrimStart('0');

It really depends on how long the NVARCHAR is, as a few of the above (especially the ones that convert through IntXX) methods will not work for:

String s = "005780327584329067506780657065786378061754654532164953264952469215462934562914562194562149516249516294563219437859043758430587066748932647329814687194673219673294677438907385032758065763278963247982360675680570678407806473296472036454612945621946";

Something like this would

String s ="0000058757843950000120465875468465874567456745674000004000".TrimStart(new Char[] { '0' } );
// s = "58757843950000120465875468465874567456745674000004000"

Code to avoid returning an empty string ( when input is like "00000").

string myStr = "00012345";
myStr = myStr.TrimStart('0');
myStr = myStr.Length > 0 ? myStr : "0";

return numberString.TrimStart('0');

TryParse works if your number is less than Int32.MaxValue. This also gives you the opportunity to handle badly formatted strings. Works the same for Int64.MaxValue and Int64.TryParse.

int number;
if(Int32.TryParse(nvarchar, out number))
{
   // etc...
   number.ToString();
}

Using the following will return a single 0 when input is all 0.

string s = "0000000"
s = int.Parse(s).ToString();

This Regex let you avoid wrong result with digits which consits only from zeroes "0000" and work on digits of any length:

using System.Text.RegularExpressions;

/*
00123 => 123
00000 => 0
00000a => 0a
00001a => 1a
00001a => 1a
0000132423423424565443546546356546454654633333a => 132423423424565443546546356546454654633333a
*/

Regex removeLeadingZeroesReg = new Regex(@"^0+(?=\d)");
var strs = new string[]
{
    "00123",
    "00000",
    "00000a",
    "00001a",
    "00001a",
    "0000132423423424565443546546356546454654633333a",
};
foreach (string str in strs)
{
    Debug.Print(string.Format("{0} => {1}", str, removeLeadingZeroesReg.Replace(str, "")));
}

And this regex will remove leading zeroes anywhere inside string:

new Regex(@"(?<!\d)0+(?=\d)");
//  "0000123432 d=0 p=002 3?0574 m=600"
//     => "123432 d=0 p=2 3?574 m=600"