BASH: let output like 00004 and not 4

bglnelissen · 09-27-2005, 07:20 AM

Well, i downloaded largefile.tar.bz2 but is was corrupted. So i did a bzip2recover and now i have allot of small .tar files. way too many to handle at ones. (when trying to "tar xvf *" it says: too many arguments (translated)).
So i wanted to write a little program that handles the files one by one. But i ran into a little problem.

The files are like:
rec00001file.tar
rec00002file.tar
rec00003file.tar
etc etc etc

i thought of a while loop and add 1 on the file name every loop.
like:
tar xvf rec00001file.tar;
tar xvf rec00002file.tar;
tar xvf rec00003file.tar;

but here is my problem. the let statement has such an output:

00001 becomes 1
00493 becomes 493
etc

(TEST IT YOURSELVES IN A CONSOLE)

Code:

let x=00000004;echo $x

but i want 00024 to stay 00024, how can i change this? is there an extra command for the statement let to do this?

thnx,
bas nelissen

(i read something about $i, but dont understand it)

Code:

#!/bin/bash
#tar-multi-unpacker
#made by b.nelissen

#QUESTIONS:
echo "hello world"
echo "starting you unpacking"
echo "whats the variable number of your first unpack item?:"
echo "     (like: 0001   is in this file:   rec0001file.tar)"
read number
echo "I am using:  $number"
echo ""

echo "whats the variable number of your last unpack item?:"
echo "     (like: 0845   is in this file:   rec0845file.tar)"
read numberhigh
echo "I am using:  $numberhigh"
echo ""

echo "whats the filename called BEFORE the variable number?:"
echo "     (like: rec    is in this file:   rec0845file.tar"
read befornumber
echo "I am using:  $befornumber"
echo ""

echo "whats the filename called AFTER the variable number?:"
echo "     (like: file.bz2    is in this file:   rec0845file.tar"
echo "     (its possible and mightbe easier to use a wildcard like: * )"
read afternumber
echo "I am using:  $afternumber"
echo ""

echo "where do you want me to unpack it to?"
echo "     (like: /home/bla/bla)"
echo "     (keep it empty for unpacking in the same dir)"
read outputdir

echo ""
echo "this is gonna be my command:"
echo " tar xvf $befornumber$number$afternumber $outputdir"
echo "to cancel press:  Ctrl-c"
sleep 10

#EXAMPLE
#while [ $number -le $numberhigh ];
#do echo $number;
#let number=$number+1;
#echo "";
#done


while [ $number -le $numberhigh ];
	echo $number;
	do tar xvf $befornumber$number$afternumber $outputdir;
        let number=$number+1;
        echo "";
        done

druuna · 09-27-2005, 10:54 AM

Hi,

The 0's (zero's) will be stripped, but you can pad variables.

Maybe you can use parts of this example (only echo's to terminal, nothing is actually done. based very loosly on the bottom part of your example).

Code:

#!/bin/bash

number=0
numberhigh=5
befornumber="before"
afternumber="after"
outputdir="/tmp/"

while (( $number <= $numberhigh ))
do
  paddednumber="`printf "%06d" $number`"
  echo "tar xvf ${befornumber}${paddednumber}${afternumber}.tar ${outputdir}"
  let number=number+1
  echo ""
done

This paddednumber="`printf "%06d" $number`" being the important part. The 06 part tells printf to add 0 (zero's) up to and including 6 chars.

For details on formatted printing: man 3 printf

Hope this gets you going again.

AnanthaP · 09-27-2005, 10:59 AM

for i in (rec*tar)
do
echo $i
done

Something like that???

Dark_Helmet · 09-27-2005, 11:05 AM

Might I suggest:

Code:

find . -maxdepth 1 -iname "rec*file.tar" -exec tar xvf {} \;

wmakowski · 09-27-2005, 03:11 PM

Similar to druuna's, but using a for loop.

Code:

#!/bin/bash
#
# Increment number to become part of file name.
#
                                                                                
for (( i=1; i<100; i++ ));do
  filenum=`printf "%05d" $i`
  fname=rec${filenum}file.tar
  echo $fname
done

Bill

bglnelissen · 09-27-2005, 04:07 PM

thnx for clearing this out.

a little test run with 13 as lucky numer:

Code:

$  paddednumber="`printf "%06d" 13`";echo $paddednumber

000013

So it works like a charm.

eddiebaby1023 · 09-27-2005, 05:02 PM

Quote:

Originally posted by bglnelissen
thnx for clearing this out.

a little test run with 13 as lucky numer:

Code:

$ paddednumber="`printf "%06d" 13`";echo $paddednumber 000013

So it works like a charm.

I'd recommend using $(..) instead of backticks - much easier to nest and looks neater, too:

Code:

$  paddednumber="$(printf "%06d" 13)";echo $paddednumber

bigearsbilly · 09-28-2005, 08:36 AM

It's dead easy this one

Code:

$ typeset -Z4 x
$ x=2
$ echo $x
0002
$ x=123
$ echo $x
0123
$ x=9  
$ echo $x
0009

wmakowski · 09-28-2005, 09:16 AM

typeset -Z works well in ksh, but not so good in bash.

$ typeset -Z5 x=2
bash: typeset: -Z: invalid option
typeset: usage: typeset [-afFirtx] [-p] name[=value] ...

bigearsbilly · 09-28-2005, 09:24 AM

doh

I forgot that one!

well, IMHO another example of how bash is far inferior for scripting.

bash a bit better interactively though I guess.