5 ms·
Point being that "ls /path; rm $!" deletes something entirely different than "/path". What you want is "ls /path; rm $_". Even then, the above is fairly point
by BossingAround 5y ago
Point being that "ls /path; rm $!" deletes something entirely different than "/path".
What you want is "ls /path; rm $_".
Even then, the above is fairly pointless. At the time you look at what you are deleting, it's gone.
- witrak 5y agoI think you misunderstood the previous comment. I think that firstly "ls /path" is entered and if the result is ok, then "; rm $_" is added to the copy of the command then executed.
- BossingAround 5y agoYes, at which point you simply do: 1. ls /path 2. rm <esc><.> (where "escape" "dot" brings up last argument) without the need to fiddle around with dollar signs underscores, exclamation marks, etc. to prevent further mistakes (e.g. "was it $! or !$ ?", shell expansion, etc)