Calling shell functions with xargs

问题

I am trying to use xargs to call a more complex function in parallel.

#!/bin/bash
echo_var(){
    echo $1
    return 0
}
seq -f "n%04g" 1 100 |xargs -n 1 -P 10 -i echo_var {} 
exit 0

This returns the error

xargs: echo_var: No such file or directory

Any ideas on how I can use xargs to accomplish this, or any other solution(s) would be welcome.

回答1:

Exporting the function should do it (untested):

export -f echo_var
seq -f "n%04g" 1 100 | xargs -n 1 -P 10 -I {} bash -c 'echo_var "$@"' _ {}

You can use the builtin printf instead of the external seq:

printf "n%04g\n" {1..100} | xargs -n 1 -P 10 -I {} bash -c 'echo_var "$@"' _ {}

Also, using return 0 and exit 0 like that masks any error value that might be produced by the command preceding it. Also, if there's no error, it's the default and thus somewhat redundant.

@phobic mentions that the Bash command could be simplified to

bash -c 'echo_var "{}"'

moving the {} directly inside it. But it's vulnerable to command injection as pointed out by @Sasha.

Here is an example why you should not use the embedded format:

$ echo '$(date)' | xargs -I {} bash -c 'echo_var "{}"'
Sun Aug 18 11:56:45 CDT 2019

Another example of why not:

echo '\"; date\"' | xargs -I {} bash -c 'echo_var "{}"'

This is what is output using the safe format:

$ echo '$(date)' | xargs -I {} bash -c 'echo_var "$@"' _ {}
$(date)

This is comparable to using parameterized SQL queries to avoid injection.

I'm using date in a command substitution or in escaped quotes here instead of the rm command used in Sasha's comment since it's non-destructive.

回答2:

Using GNU Parallel is looks like this:

#!/bin/bash
echo_var(){
    echo $1
    return 0
}
export -f echo_var
seq -f "n%04g" 1 100 | parallel -P 10 echo_var {} 
exit 0

If you use version 20170822 you do not even have to export -f as long as you have run this:

. `which env_parallel.bash`
seq -f "n%04g" 1 100 | env_parallel -P 10 echo_var {}

回答3:

Something like this should work also:

function testing() { sleep $1 ; }
echo {1..10} | xargs -n 1 | xargs -I@ -P4 bash -c "$(declare -f testing) ; testing @ ; echo @ "

回答4:

Maybe this is bad practice, but you if you are defining functions in a .bashrc or other script, you can wrap the file or at least the function definitions with a setting of allexport:

set -o allexport

function funcy_town {
  echo 'this is a function'
}
function func_rock {
  echo 'this is a function, but different'
}
function cyber_func {
  echo 'this function does important things'
}
function the_man_from_funcle {
  echo 'not gonna lie'
}
function funcle_wiggly {
  echo 'at this point I\'m doing it for the funny names'
}
function extreme_function {
  echo 'goodbye'
}

set +o allexport

来源：https://stackoverflow.com/questions/11003418/calling-shell-functions-with-xargs

标签

bash

xargs