Appending file extensions to filenames in bash scripts

Applications UNIX Desktop Questions & Answers
1

Hi

Suppose I have a variable called filename and it it contains the name of a file. I then would like to append an extension to that filename. The filename currently has no extensions.
How do I do this?

Thanks

2

Say you wanna add a .txt extension to the variable

mv $filename ${filename}.txt
3

Hi

Thanks for your reply. It looks like a relatively easy task, however, it doesn't work on my side.

Here is my code:
First I remove the old file extension

filename=${cfile%%.*}

Then I append the new extension

 mv $filename $filename.dsx

And echo the filename to see that our new file has the correct name

echo ${filename##*/}

This is my output:

$ ./Import_PlannedBSC.sh -b
mv: cannot stat `/var/local/dsx/csv/ds_pl_bsc_tester': No such file or directory
ds_pl_bsc_tester
mv: cannot stat `/var/local/dsx/csv/rnChk_pos_bsc_orig_001': No such file or directory
rnChk_pos_bsc_orig_001
4

Hi.

Shouldn't that be

 mv $cfile $filename.dsx
5

Hi Scottn

This methos gives me the following output:

$ ./Import_PlannedBSC.sh -b
ds_pl_bsc_tester
rnChk_pos_bsc_orig_001

You see, this gives me filenames without the .dsx file extension that I desperately need.

I was looking at the renaming convention as well, still reading up on how to possibly make that work.

6

I guess you echo at the end the variable "filename" so it prints the without extension which is correct.

7

If you can post the complete code it will be easier to debug.

8

Okay it's quite long so here goes:

I have files stored in different directories, hence I have put directory names into variables:

dsxfile="$dsxfile"
csvfile="$csvfile"
filename=""
cfile=""
csvdir="/var/local/dsx/csv"
dsxdir="/var/local/dsx/import"

Here is my scriptname that is called like so from the command line:

./Import_PlannedBSC.sh -b

I have getops that when I call the script with the -b option, it calls a function batchConverter which is the most important part here

while getopts " c: d: b h " option
do
	case $option in
		c ) csvfile="$OPTARG";;
		d ) dsxfile="$OPTARG";;
		b ) batchConverter;exit;;
		h | ? | * ) displayHelp;exit;;
		
	esac;
done

Here is my function which pulls all files that have filename "_bsc_" pattern and then removes it's file extension called .csv and then appends a new file extension called .dsx

function batchConverter
{
	for cfile in $csvdir/*_bsc_*;
	do
		filename=${cfile%%.*} 
		 mv $cfile $filename.dsx  
		echo ${filename##*/} 
  	      
	done
}

I want the filename without the path. The I copy these new files to another directory called $dsxdir

9 
filename=${cfile%%.*}.dsx
mv $cfile $dsxdir/${filename##*/}
1 Like
10

At what point do I add the new file extension?

11

You can do

filename=${cfile%%.*}.dsx
mv $cfile $dsxdir/${filename##*/}

Or

filename=${cfile%%.*}
mv $cfile $dsxdir/${filename##*/}.dsx

It really makes no difference...

12

Thank you, the second option works brilliantly. Problem is that both the files are moved to the $dsxdir directory, but it's no problem, I will try and sort that out.
Thank you so so so much for all your help