bash script: how do I refer to a counter i in a "for i in" loop

frankie_DJ · 10-01-2006, 04:16 PM

Hi,

Is it possible to refer to counter i in a
for i in *
loop? How do I do that?

I just wanted to make my file names shorter after processing. Like I have files
anneal_i.radial_Si_Si
where i goes from 1 to 20, and I want to generate some other files with a script, but also in the process reduce names of the ones I have. Thanks

Mara · 10-01-2006, 04:32 PM

It's $i. Example:

Code:

for i in 1 2 3 ;do echo file_$i; done

The result:

Code:

file_1
file_2
file_3

frankie_DJ · 10-01-2006, 04:51 PM

I know that. I want to refer to i itself. like for i in file[1-6]
I want to refer to 3, not file3.

$3 refers to file3 not 3. what refers to 3?

demon_vox · 10-01-2006, 05:10 PM

Hi,
the for loop iterates over the elements you provide them. As far as bash is concerned all the elements are strings. They dont have the format you want and are magical parse so that something automatically point at the part you want.
$i will always point at the current element. If you want part of that element, then you have to parse it yourself.

Hope this is useful.
Cheers!

/bin/bash · 10-02-2006, 03:22 AM

Quote:

I know that. I want to refer to i itself. like for i in file[1-6]
I want to refer to 3, not file3.

$3 refers to file3 not 3. what refers to 3?

$for i in `seq 1 20`;do echo "This is loop $i";done
This is loop 1
This is loop 2
This is loop 3
...
This is loop 20

Then to rename a file do this:
$for i in `seq 1 20`;do mv anneal_$i.radial_Si_Si anneal_$i.radial;done

gloriant · 07-24-2007, 08:45 AM

using

Code:

$( seq <start> <stop> )

is one way to do it. By the way, i prefer the $( ... ) notation over the back quote, because sometimes back quotes are incorrectly interpreted/read/remembered/copied by readers.

Sometimes you'll want to allow the loop-code to adjust the counter, which then asks for something like:

Code:

for (( counter=0; ${counter} < 20; counter=$(( ${counter}+1 )) )); do
  echo "this is loop ${counter}.";
done

in this code, mind the separating spaces around (( and )).

Or you do it completely by hand:

Code:

counter=0;
while [ ${counter} -lt 20 ]; do
  echo "this is loop ${counter}.";
  counter=$(( ${counter}+1 ));
done

OR, if you have some files, but want to get the number that's in the file, I'd do it like this:

Code:

for file in file[0-9]*; do
  i=$(echo "${file}"|sed 's/^file//;');
  echo "file index ${i}";
done

so decrease all files by one for instance:

Code:

for file in file[0-9]*; do
  i=$(echo "${file}"|sed 's/^file//;');
  i=$(( ${i} - 1 ));
  if [ ! -f file${i} ]; then
    mv ${file} file${i};
  else
    echo "file${i} exists already.";
  fi
done

or then, when increasing files:

Code:

function increase_file {
  local newfile;
  fn=increase_file;
  if [ 1 != ${#} ]; then echo "${fn} invalid nr of arguments ${#}.";
  elif [ ! -f ${1} ]; then echo "${fn} called with non-file argument.";
  else
    i=$(echo "${1}"|sed 's/^file//;');
    i=$(( ${i}+1 ));
    newfile=file${i};
    if [ -f ${newfile} ]; then
      echo "first increase existing file ${newfile}.";
      increase_file ${newfile};
    fi
    mv ${1} ${newfile};
  fi
}

# calling with:
increase_file file1;