passing multiple bash commands to a function-CodePudding

I want to pass in multiple commands using a && operator into a single function call. Each command could have a different number of parameters so I'm using the $@ to capture all of them. With a single command, the function below works fine.

function try_log () {
     $@ \
         && echo "PASS!"
         || echo "FAIL!"
}

$ try_log touch foo bar
  PASS!
$ try_log rm foo bar
  PASS!
$ try_log rm foo bar
rm: cannot remove 'foo': No such file or directory
rm: cannot remove 'bar': No such file or directory
  FAIL!

I now want to add support for an && operatoron the input command. The intent is to run the entire command then display a PASS or FAIL at the end. I've tried the following without success.

try_log ls && ls # runs the send ls after the first completes
foo bar
  PASS!
foo bar

try_log 'ls && ls'
ls: cannot access '&&': No such file or directory
ls: cannot access 'ls': No such file or directory
  FAIL!

$ try_log `ls && ls`
foo: command not found
  FAIL!

CodePudding user response：

First, your function does not work fine. You need to quote $@ to handle arbitrary arguments, and you should just use a regular if statement rather than trying to simulate a conditional expression using && and ||.

try_log () {
     if "$@"; then echo "$PASS!"; else echo "FAIL!"; fi
}

The problem is that this only handles simple commands, consisting of the name of a command and its arguments. It does not handle pipelines, compound commands like [[, if, etc, or command lists like a && b. The only real solution is to pass the command as a single string and let try_log use eval to parse and execute it.

try_log () {
     if eval "$1"; then echo "$PASS!"; else echo "FAIL!"; fi
}

try_log "ls && ls"

(This is not an "evil" use of eval, because you are using it exactly what it is intended for: executing arbitrary code. Just make sure you, and not an untrusted user, are providing the arbitrary code.)