#!/bin/bash
function func_name {
arg1=$1; arg2=$2;
if [[ "${arg1}" == "abcd" ]]; then
echo enterd abcd condition
else
echo entered else
fi
}
export -f func_name
find . -name "*" -type d -exec bash -c '( cd {} &&
func_name;
)' bash $1 $2 {} \;
Trying to run a function having if condition as a subshell inside find statement. It just enters else part. Why?
It's not clear what arguments should be passed where, but your current attempt is certainly wrong. You aren't passing any arguments to func_name when you are calling it, so "$arg1" is expanding to the empty string. I suspect you want something like
find . -name "*" -type d -exec bash -c 'cd "$1" && func_name "$2" "$3"' {} "$1" "$2" \;
Here, func_name is explicitly given the arguments passed to the original script. You don't need the subshell in the argument to -c, since the entire command is already run in a separate process.
If you are using bash 4 or later, you probably don't actually need to use find.
shopt -s globstar
for d in **/*/; do
pushd "$d"
func_name "$1" "$2"
pops
done
Variables inside a function block are only evaluated when the function is called. If you want to bind some variables at function declaration time, use eval, or a pair of global variables.
#!/bin/bash
_private_foo=$1
_private_bar=$2
func_name {
if [[ "$_private_foo" == "abcd" ]]; then
echo entered abcd condition
else
echo entered else
fi
}
If subsequent code never changes the global variables _private_foo and _private_bar, the function will work as if the initial global values of $1 and $2 were interpolated into the function definition, without the myriad complications and perils of eval.
If you need the function to work in a subshell, you'll need to also export the globals, or somehow otherwise pass them in, as pointed out by @chepner.
