Print a variable while it's doing the action?

Dan8080 · 03-07-2009, 05:51 PM

Hi,

I'm pretty new to bash programming and I'm stuck on something. I don't even know if there's a way to do this or if I'm in the wrong direction here. Example script:

Code:

#!/bin/bash
files=`ls`;

How can I store that command in a variable like I have there and at the SAME TIME get it to print itself while it's doing it? If more information is needed about what I'm actually trying to make, I can provide it.

Thanks in advance.

unSpawn · 03-07-2009, 06:55 PM

Do you mean something like:

Code:

cmd=/bin/ls
function cmd() { echo "$FUNCNAME: $cmd" > /dev/stderr; $cmd; }
files=$(cmd)

Quote:

Originally Posted by Dan8080

If more information is needed about what I'm actually trying to make, I can provide it.

It never hurts to clarify what you're trying to accomplish in an OP. Faster and more efficient for everyone IMHO.

Hko · 03-08-2009, 04:21 AM

Or run the script with the "-x" option.

Code:

#!/bin/bash
set -x  # Switch on printing commands as they are executed
files=`ls`;
set +x  # switch it off again

Or, for one time only, run the script this way:

Code:

bash$ bash -x /path/to/script.sh

David the H. · 03-08-2009, 06:31 AM

I understand exactly what he's asking. He wants the output of the embedded command (ls in this case) to simultaneously be stored in the variable and echo to the standard out (i.e. display normally). And now I'm curious if this can be done myself.

But is there any reason why you can't simply put an "echo $files" line in the script right after setting the variable, so that you can see the output right away?

Code:

files=$(ls)
echo $files

Hko · 03-08-2009, 07:21 AM

Is is possible to print the output of the command to stderr at the same time.

Code:

lstmp=$(ls /tmp | tee >(cat >&2))
echo $lstmp

But does i make sense? For "debugging" I'd prefer the -x option. For other purposes you can just "echo $lstmp" like you said.

Dan8080 · 03-08-2009, 12:32 PM

Thanks for the help everyone. I'll try to provide more info to explain why I want this.

Here's a piece from my backup script that uses rsync:

Code:

echo "               [ Files are being backed up.]                        "
echo "===================================================================="
fbkres=`rsync --bwlimit=6000 -a -v ${www_path} ${ext_hd_path}Server/html/ --stats --progress --exclude=mysql --exclude=gamedata`

echo "===> Writing log file of results..."

wrtolfl= `echo "===================================== " >> ${ext_hd_path}backup_logs/file-backup-results.txt`
wrtolfl= `echo "File Backup " >> ${ext_hd_path}backup_logs/file-backup-results.txt`
wrtolfl= `echo "Date: $(date +%Y) / $(date +%m) / $(date +%d)" >> ${ext_hd_path}backup_logs/file-backup-results.txt`

# Advanced logging removed
#wrtolfl= `echo "" >> ${ext_hd_path}backup_logs/file-backup-results.txt`
#wrtolfl= `echo "Results: $fbkres" >> ${ext_hd_path}backup_logs/file-backup-results.txt`
#wrtolfl= `echo "" >> ${ext_hd_path}backup_logs/file-backup-results.txt`

wrtolfl= `echo "===================================== " >> ${ext_hd_path}backup_logs/file-backup-results.txt`
wrtolfl= `echo "" >> ${ext_hd_path}backup_logs/file-backup-results.txt`

echo "===> Done..."

Here's the problem - I added the "--progress" because I'd like to actually see this while it runs, but if I get rid of it in the variable then I can't make a log of the stats that rsync outputs at the end as I have been doing. I can't echo it AFTER because there's no point there; I want to see the backup progress live. That's why I'm wondering if there's a way to get variables to "echo while they run." I was planning to just tail the variable and insert the stats that way.

I think Hko's solution to use set -x is exactly what I need, I will test when I am able to - thanks!

Hko · 03-08-2009, 02:05 PM

Ah! Now it makes sense to me why you ask this.

You want to store the output of rsync in a variable so you can store the output in a log later. But you also want to see the output as it runs, right?

Though "set -x" would more or less do it for you, I think it is not the best solution. In fact, storing the output in a variable and the later echo it to a file is not a very good idea IMO. If many files are backed up the variable may consume quite a lot of memory (the --progress option will make rsync write two lines for every file backed up). Also, the echo command will write it to the file as one single big line (newline-codes are lost).

There is a simple solution to solve all this: use the "tee" program.

Instead of what you are trying to do:

Code:

logvar=`rsync -av --stats --progress /srcpath /destpath`
# ...
echo "$logvar" >/path/to/logfile

You can do this to write to a file and stdout (the terminal) at the same time:

Code:

rsync -av --stats --progress /srcpath /destpath | tee /path/to/logfile