CODEWITHSUNDEEP
bashfunctionshellloopsfind
lenzinho
lenzinho

Reputation: 401

pass results from find to a custom bash function?

I have a custom bash test function

main () {
    list=$1
    for i in ${list}; do
        echo $i
    done
}

which I am trying to call as:

main `find . -name hello`

but it is only echoing the first file passed in the list. The original output of find . -name hello is

./world1/hello
./world2/hello
./world3/hello
./world4/hello
./world5/hello
./world6/hello
./world7/hello
./world8/hello

How can I make sure the whole list is passed?

Thanks!

Upvotes: 1

Views: 395

Answers (1)

Inian
Inian

Reputation: 85800

Use a proper Process-Substitution syntax to loop over files, and you don't need a function if you are looking to do a certain action on each of the files from find output, just do

#!/bin/bash

while IFS= read -r file
do
    echo "$file"
    # Your other actions on "$file" go here
done < <(find . -type f -name "hello")

(or) if you are afraid of having special characters in file-names like a space or any of the meta-characters, use the -print0 option if it is supported (GNU findutils package) to separate the files on \0 de-limiter and read it back with the same de-limiter.

while IFS= read -r -d '' file
do
    echo "$file"
    # Your other actions on "$file" go here
done < <(find . -type f -name "hello" -print0)

I just simply do NOT recommend using function this-way, but you can do something like,

function findRead() {   
    while IFS= read -r file
    do
        echo "$file"
        # Your other actions on "$file" go here
     done <<<"$1"    
}

findRead "$(find . -type f -name "hello")"

(or) more-simply just do,

function findRead() {    
    for file in "$1"
    do
        echo "$file"
    done
}

findRead "$(find . -type f -name "hello")"

The key here is to double-quote the $(..) command-substitution that returns the output of find, to preserve the new-line characters which are then fed to your custom function.

Upvotes: 2

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