Question

I write a bash script:

   #!/bin/bash
    function test_echo
    {
        echo $0
        echo $1
        echo $2
        echo $#
    }
    test_echo

I try:

find test.sh -type f -exec test_echo '{}' \;

or

find . -type f -exec `./test.sh {}` \;

but it doesn’t work.

I need a way to scan file from a folder (I use find) and for each file call a function ( foo_fonction() ) with in parameter the full path of the file.

 find . -type f -exec foo_fonction '{}' \; 

And how to have the parameter (full path) in the foo_fonction() ?

Answer 1

There are 2 ways to go about this:

1: You can put either put the function definition as part of your find files script:

(Let’s call this find_stuff.sh)

#!/bin/bash

test_echo() {
    echo $0
    echo $1
    echo $2
    echo $#
}

files=$(find $(pwd) -type f)
for f in $files
do
    test_echo "$f"
done

Explanation of what this does:

• A new shell is spawned to run this script if you do ./find_stuff.sh
• First it defines the test_echo() function,
• Then it finds all of the files under the current working directory recursively
• Note that since we specified $(pwd) as the search directory, which will be an absolute path, the results returned by find will be absolute paths too – as you desired.
• Then for each of these files, pass its file path to the test_echo function

2: Put the function definition elsewhere, but you still want a find_stuff.sh script:

#!/bin/bash

files=$(find $(pwd) -type f)
for f in $files
do
    ./test.sh "$f"
done

and test.sh should look like this:

#!/bin/bash

test_echo () {
    echo $0
    echo $1
    echo $2
    echo $#
}

test_echo $@

Note the last line test_echo $@ – the $@ part is what you were missing before.