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bashsedquartz-composer

SED, Getting the last file name

It is a SED issue.

I am working in Quartz Composer (Mac OSX) script to:

  1. Open the last .log from a directory.
  2. Getting the last text line of that file and remove its 24 first characters.
  3. Show the last line each time a new one is created inside the last .log

I think this would be so simple. At this moment, I have this done in QC, and it is almost working. But I have issues with the SED commands in both getFileName.sh and getLastLineOfLog.sh

getFileName.sh

#!/bin/bash

cd /Users/Me/
/bin/echo `ls -tr *.log | tail -1`

What I need here is to open the last .log file, with this name structure 127.0.0.1,NUMBER.log being the NUMBER section a 5 digits number. The higher the number, the newer the .log file. Now, it opens just the first .log file.

Do you know how to write in SED the right commands for doing that, please?

getLastLineOfLog.sh

#!/bin/bash

/usr/bin/tail -1 /Users/Me/$1 | sed 's/^.\{24\}//' > /Users/Me/$1.log
echo `/usr/bin/tail -1 /Users/Me/$1.log`

In this second file, I have tried to get the last file of the loaded .log and echo it without its 24 first characters. It works, but creating a new .log.log. (etc.) file each time the function passes, thing that I would like to avoid.

Is it necessary to output the results of the SED commands to a new file? How could it be done as simple as possible?

Thank you in advance.

Solution

  • if you want to sort by the filename:

    ls *.log | sort -t, -k2 -n | tail -1