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I'm somewhat new to shellscripting. I am attempting to make a shellscript that when exececuted, will execute the second to last and the last executed commands in my terminal in that order. If I go to my terminal and type !! the last command will be executed. Similarly, if I type !-2 the second-to-last command will be executed. I am effectively trying to make a shellscript that when executed will perform the equivalent of me typing !-2 into my terminal twice in succession. I've tried numerous ideas (echo !-2, !-2, etc.) and honestly can't figure out how to get this to work. Can anyone give me any pointers?

supersmarty1234
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  • Does this answer your question? https://unix.stackexchange.com/a/5701/411962 – fuzzydrawrings Feb 17 '22 at 06:37
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    You want to execute the second to last command, in _what_ shell's history? Should this work even when called from e.g. a cron job? – Kusalananda Feb 17 '22 at 06:54
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    You probably want to do this as a function, not as a script. A script runs in a new shell which doesn't inherit the current shell's in-memory history. At most, you can get it to load in your shell's history **file** (e.g. ~/.bash_history), or extract the last two lines of that file with `tail`. The history file probably won't have whatever commands you've run in the current shell session, only whatever was most recently saved to it. – cas Feb 17 '22 at 09:46
  • @fuzzydrawings Not quite, I haven't been able to get something working yet, but it was a helpful read, I'm trying something now along the lines of what cas suggested. Extracting the last three lines should actually work for my purposes, and then it'd just be a matter of executing the -3rd and -2nd prior commands, since the last run command would be the script itself. – supersmarty1234 Feb 17 '22 at 17:15

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