CODEWITHSUNDEEP
bashshell
Gary
Gary

Reputation: 997

Capturing and storing a value from the output of a shell script while also displaying progress?

I use a bash script from an upstream repo that echos the high level progress as it goes through the script. At the end it displays a path/filename for the results of the script. I'm trying to get that path/file name without modifying the script since the script changes frequently. I can't figure out how to get the path/filename.

I've spent a lot of time and thought now might be a good time to ask:

How should I do this?

Here's a sample just to demonstrate the output.

#!/bin/bash

echo "The directory to be analyzed abc123/def456/"
echo "## CPU"
echo "## Messages"
echo "## Out of Memory"
echo "Calling other script..."
echo "Done."
echo
echo "## Please check out the file /tmp/report_user1/def456-2020-12-02.log"

Output. I'm trying to grab the path/filename from the last line.

The directory to be analyzed abc123/def456/
## CPU
## Messages
## Out of Memory
Calling other script...
Done.

## Please check out the file /tmp/report_user1/def456-2020-12-02.log

Thank-you.

Upvotes: 1

Views: 47

Answers (1)

Milag
Milag

Reputation: 1986

To view stdout on your terminal plus capture / filter, one method without tmp files is piping the output from any [utility or script] into tee /dev/tty, along with piping the tee output to a filter, eg:

util | tee /dev/tty | grep '\.log$'

capturing the above to a bash variable:

var=$(util | tee /dev/tty | grep '\.log$')

From the OP's description the above could be scripted, tee output could reduced by other methods, log file copied, then exec $EDITOR ./copy.log

Upvotes: 1

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