Remove leading zeros in batch file

Question

In my application, I get a number having leading zeros. I am trying to trim the leading zeros and get the actual number. I tried using /a switch which considers right side of the assignment as an arithmetic expression. So I tried:

SET /a N = 00027 

The above gave me the output of 23 which is the decimal equivalent of octal number 27. Then I found this solution online.

SET N = 00027 SET /a N = 1%N%-(11%N%-1%N%)/10 

This seems working and is giving the output 27. Is there much easier way to trim the leading zeros in a batch file?

Answer 1

The method you found is very good. It supports numbers up to 99,999,999 and is very fast.

There is a simpler way to use SET /A that works with numbers up to 9999. It uses the modulus operation. The method cannot be extended to larger numbers.

 set n=0027  set /a n=10000%n% %% 10000 

The FOR /F method that Dale posted works with "any" size number (up to 8191 digits). However, it needs just a bit more work to handle zero values.

set n=000000000000000000000000000000000000000000000000000027 for /f "tokens=* delims=0" %%N in ("%n%") do set "n=%%N" if not defined n set "n=0" 

Answer 2

The exit command is pretty good at clearing leading zeros:

>set n=0000890  >cmd /c exit /b %n%  >echo %errorlevel% 890 

With this you can use number up to 32 bit integer limit and the piece of code is really small, despite additional call of cmd.exe could harm the performance.