Reputation: 3001
I am trying to run many commands in parallel, and need to do some string manipulation on the input first. How can I make the below example work?
find . -mindepth 1 -type d | xargs -n 1 -P 20 -i sh -c "v={}; echo $v"
When I use this, $v
is null. Why is it not being saved as the value of {}
?
Upvotes: 10
Views: 19351
Reputation: 33685
With GNU Parallel you would do:
find . -mindepth 1 -type d | parallel -P 20 'v={}; echo "$v"'
Upvotes: 1
Reputation: 4612
The parent shell is expanding $v
before the string gets passed to xargs
.
Suppose your find
command finds a subdirectory named ./stuff
.
First, the parent bash
shell (the one you typed the find
command into) will expand $v
, because the string is in double quotes. You currently have no value set for variable v
, so it expands to an empty string.
Next, the arguments get passed to xargs
, which will see this: v={}; echo
Then, xargs
will read ./stuff
from the pipe, and replace {}
with ./stuff
Finally, the sh
command is executed by xargs
, and sh
will see this: v=./stuff; echo
To fix this, you need to either escape the $
so that the parent shell doesn't expand it, or use single quotes to avoid variable expansion. You should also probably quote the strings so that any directory names with spaces in them don't cause problems with the final sh
command:
find . -mindepth 1 -type d | xargs -n 1 -P 20 -i sh -c "v=\"{}\"; echo \"\$v\""
OR
find . -mindepth 1 -type d | xargs -n 1 -P 20 -i sh -c 'v="{}"; echo "$v"'
With either command, the final sh
process will see: v="./stuff"; echo "$v"
By the way, one way to see for yourself that this is indeed what is happening would be to set a value for v
in the parent shell, then run your original command. The shell will expand $v
to whatever value you set, and you will see that value repeated for every directory found by find
.
$ v=foobar
$ find . -mindepth 1 -type d | xargs -n 1 -P 20 -i sh -c "v={}; echo $v"
foobar
foobar
foobar
foobar
foobar
foobar
foobar
foobar
foobar
foobar
foobar
...
Upvotes: 14