On Sun, May 4, 2014 at 4:32 PM, ~Stack~ i.am.st...@gmail.com wrote:
Greetings,
I know this isn't explicitly a Debian problem, but I have seen many
really smart Bash programmers on this list so I figured this was a good
place to ask. :-)
$ bash --version
GNU bash, version 4.2.37(1)-release (x86_64-pc-linux-gnu
$ cat /etc/debian_version
7.5
I have a Bash script I am working on where I am getting (supposedly) a
two-digit number from the user. I handle all the logic to ensure I am
getting two digits and not 2, letters, characters, ect in a different
function. By the time I get to this code in my script, I am assuming it
is a one or two digit number. Because I need a two digit number I am
assuming that my user may have just input a single digit for the first
nine numbers without a leading zero. I am doing this in my code like:
verifiednum=`printf %02d $uservar`
This works really well when they enter only a single digit or 01-07.
However, on 08 or 09, this fails.
$ cat test.sh
#!/bin/bash -
printf %02d 07 #This Works.
echo
printf %02d 8 #This Works.
echo
printf %02d 08 #This doesn't.
echo
printf %02d 9 #This Works.
echo
printf %02d 09 #This doesn't.
echo
$ ./test.sh
07
08
./test.sh: line 6: printf: 08: invalid octal number
00
09
./test.sh: line 10: printf: 09: invalid octal number
00
I almost want to think that this is a bug, but because it seems to be
thinking it is an octal number (which technically it is I suppose :). I
am guessing that it just doesn't like what I am doing with printf but I
am a bit baffled as to why it only croaks on 08 and 09.
1) Does anyone know what is wrong here?
leading zero is interpreted as octal. You need to prepend the number
with 10# to convert it to decimal first
2) Is there a better way of solving this issue?
so try this instead printf %02d $(( 10#08 ))
Thanks!
~Stack~
--
Asif Iqbal
PGP Key: 0xE62693C5 KeyServer: pgp.mit.edu
A: Because it messes up the order in which people normally read text.
Q: Why is top-posting such a bad thing?
