I am trying to call a bash function and assign the output the function returns to a variable. However, when I run the code below, I get the following error and I don't understand what I may be doing wrong:
test.sh: return: line 7: Illegal number: aGVsbG93b3JsZA==
helloworld
convertfrombase64 () {
B64DATA=$1
echo "$B64DATA" | base64 -d
return "$B64DATA"
}
PLAINTEXT=$(convertfrombase64 "aGVsbG93b3JsZA==")
echo "$PLAINTEXT"
You can't return a string from a function. A Bash function's return value follows the same rules as that of exit codes - see the related post.
You are already using the output of the function correctly. May be you want to return the exit code of base64 to indicate status, this way:
convertfrombase64 () {
B64DATA=$1
echo "$B64DATA" | base64 -d
return $?
}
Based on the great suggestions got from comments, the above function could be better written as:
convertfrombase64() {
printf '%s' "$1" | base64 -d
}
See this post for more details:
return $? is actually redundant, since the default behavior is to return the exit status of the last command. In fact, the function could written as just convertfrombase64 () { base64 -d <<< "$1" ; }, which also avoids touching the global B64DATA variable, avoids any nontrivial behaviors of echo, avoids some unnecessary subshells, etc.
printf '%s' "$1" | base64 -d instead of the here string, which will add a trailing newline to the value being decoded.