How do I pass a list to for in bash?
I tried
echo "some
different
lines
" | for i ; do
echo do something with $i;
done
but that doesn't work. I also tried to find an explanation with man but there is no man for
I know, I could use while instead, but I think I once saw a solution with for where they didn't define the variable but could use it inside the loop
for iterates over a list of words, like this:
for i in word1 word2 word3; do echo "$i"; done
use a while read loop to iterate over lines:
echo "some
different
lines" | while read -r line; do echo "$line"; done
Here is some useful reading on reading lines in bash.
for where they didn't define the variable but could use it inside the loopThis might work but I don't recommend it:
echo "some
different
lines
" | for i in $(cat) ; do
...
done
$(cat) will expand everything on stdin but if one of the lines of the echo contains spaces, for will think that's two words. So it might eventually break.
If you want to process a list of words in a loop, this is better:
a=($(echo "some
different
lines
"))
for i in "${a[@]}"; do
...
done
Explanation: a=(...) declares an array. $(cmd...) expands to the output of the command. It's still vulnerable for white space but if you quote properly, this can be fixed.
"${a[@]}" expands to a correctly quoted list of elements in the array.
Note: for is a built-in command. Use help for (in bash) instead.
a="..." and for i in $a; do ... if you know, that there are no "dangerous" signs in it
a=("word" "a b" "foo") (3 elements) and then for i in "${a[@]}"; do ...; that would iterate three times. You would get 4 iterations.$a as in my example. (My intention is to lazy reformatting and piping stuff on the console to get some things done)What about this:
echo " a b c " | for i in $(xargs) ; do echo "$i"; done
Result:
a
b
c
This is not a pipe, but quite similar:
args="some
different
lines";
set -- $args;
for i ; do
echo $i;
done
cause for defaults to $@ if no in seq is given.
maybe you can shorten this a bit somehow?