Bash: Variable type confusion

[Joel Hammer]

Thanks for the pointer.
It looks like it is a BAD thing to try to use an integer as a string.
Well, I knew I missed Pascal for reason.
Joel


On Sun, Oct 20, 2002 at 10:26:17PM -0400, kwall at kurtwerks.com wrote:
> On Sun, Oct 20, 2002 at 04:53:00PM -0400, Joel Hammer wrote:
> > I have always thought that bash does the right thing when it evaluates
> > variables.
> > Until today.
> > This bit of script works as expected:
> > k=0
> > for i in `dir -1 *jpg`
> > do
> > k=$((k+1))
> > echo $k
> > done
> > 
> > It prints out the numbers from 1 to whatever the number of jpg's in the
> > directory.
> > 
> > However, this fails:
> > k=0
> > for i in `dir -1 *jpg`
> > do
> > k=$((k+1))
> > [ "$k" -eq 10 ] && k=x   <---The problem.
> > echo $k
> > done
> > 
> > Here, it prints out 1 to 9 and then every tenth item is an x, followed by 1
> > to 9 again.
> 
> Correct. [ "$k" -eq ] evaluates to true, so statement following
> the && gets executed. 
> 
> > What I am trying to do is put leading zeroes on my numbers less that 10.
> > Attempting that, with [ "$k" -lt 10 ] && k=0"$k" give a base error at 8.
> 
> Try this:
> 
> ----- cut here -----
> k=0
> for i in `dir -1 *`
> do
> 	k=$((k+1))
> 	[ $k -lt 10 ] && echo "0"$k || echo $k
> done
> ----- cut here -----
> $ ./x.sh
> 01
> 02
> 03
> 04
> 05
> 06
> 07
> 08
> 09
> 10
> 11
> 12
> 13
> 14
> 15
> $
> > Any insight appreciated.
> > 
> > Joel
> > 
> > 
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