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arraysbashparameter-expansion
fedorqui
fedorqui

Reputation: 289525

Shell parameter expansion on arrays

Say I read some data into a Bash array:

$ IFS=" " read -a arr <<< "hello/how are/you iam/fine/yeah"

Now, I want to print the first /-sliced field for each element in the array.

What I do is to loop over the elements and use shell parameter expansion to strip everything from the first /:

$ for w in "${arr[@]}"; do echo "${w%%/*}"; done
hello
are
iam

However, since printf allows us to print the whole content of the array in a single expression:

$ printf "%s\n" "${arr[@]}"
hello/how
are/you
iam/fine

... I wonder if there is a way to use the shell parameter expansion ${w%%/*} at the time of using printf, instead of looping over all the elements and doing it against every single one.

Upvotes: 7

Views: 6183

Answers (2)

Ayush Saxena
Ayush Saxena

Reputation: 1

Building on @fedorqui's answer, if OP wants to output an array instead of a string, this worked for me-

a=(alpha beta gamma)
newArr=( "${a[@]//a/f}" )

P.S. I personally arrived to this post looking to output an array after doing substitutions on each element of an array.

Upvotes: 0

fedorqui
fedorqui

Reputation: 289525

Oh, I just found the way: just use the parameter expansion normally, only that against ${arr[@]} instead of ${arr}!

$ IFS=" " read -a arr <<< "hello/how are/you iam/fine/yeah"
$ printf "%s\n" "${arr[@]%%/*}"
hello
are
iam

Greg's wiki helped here:

Parameter Expansion on Arrays

BASH arrays are remarkably flexible, because they are well integrated with the other shell expansions. Any parameter expansion that can be carried out on a scalar or individual array element can equally apply to an entire array or the set of positional parameters such that all members are expanded at once, possibly with an additional operation mapped across each element.

$ a=(alpha beta gamma)  # assign to our base array via compound assignment
$ echo "${a[@]#a}"      # chop 'a' from the beginning of every member
lpha beta gamma
$ echo "${a[@]%a}"      # from the end
alph bet gamm
$ echo "${a[@]//a/f}"   # substitution
flphf betf gfmmf

Upvotes: 10

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