CODEWITHSUNDEEP
batch-filebatch-processing
Bharath
Bharath

Reputation: 55

Batch error: Numeric constants are either decimal (17), hexadecimal (0x11), or octal (021)

I'm extracting two digits and adding one to it. But when the last two digits are 08 it's throwing the above error.

set last_two=!file_name:~-2,2!
set /a add_millisec=!last_two!+1
set add_millisec=0!add_millisec!
set add_millisec=!add_millisec:~-2!

Can someone please check and help me here...

Upvotes: 1

Views: 2507

Answers (1)

aschipfl
aschipfl

Reputation: 34949

Open a command prompt window and type set /?:

[...] Numeric values are decimal numbers, unless prefixed by 0x for hexadecimal numbers, and 0 for octal numbers. So 0x12 is the same as 18 is the same as 022. Please note that the octal notation can be confusing: 08 and 09 are not valid numbers because 8 and 9 are not valid octal digits. [...]

You will find that set /A treats numbers with leading 0 as octal numbers.

To overcome this, you can do the following:

  1. Prefix the number with 1 and remove it again after the calculations:

    set last_two=1!file_name:~-2,2!
set /A add_millisec=last_two+1
set add_millisec=!add_millisec:~-2!

    For this you need to know the total number of digits in advance.

  2. Remove the trailing zeros before any calculations and pad the result with leading zeros afterwards as needed:

    set last_two=!file_name:~-2,2!
rem The following line removes all leading zeros:
for /F "tokens=* delims=0" %%Z in ("!last_two!") do set last_two=%%Z
set /A last_two+=0 & rem this avoids `last_two` to be empty if it is `0`
set /A add_millisec=last_two+1
set add_millisec=0!add_millisec!
set add_millisec=!add_millisec:~-2!

    Or:

    set last_two=!file_name:~-2,2!
rem The following two lines remove all leading zeros:
cmd /C exit !last_two!
set /A last_two=!ErrorLevel!
set /A add_millisec=last_two+1
set add_millisec=0!add_millisec!
set add_millisec=!add_millisec:~-2!

    This method is more flexible as you do not know the number of digits.

Upvotes: 3

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