This is my function
validateFile()
{
echo "$1" | grep '.zip$' > /dev/null
if [ $? -eq 0 ]
then
return 1
else
return 0
fi
}
printf "\n Enter Source File1 Path: "
read file1
res=$(validateFile $file1);
echo "$res"
But nothing store in **"res"** store nothig
validateFile function verify that file is zip file or not if yes then return 2 else 0.And returned value stored in res variable. But issue is no value store in res variable.
Although shell script provides a return statement, the only thing you can return with, is the function's own exit status (a value between 0 and 255, 0 meaning "success").
Best way to return a value is to echo the same, something like below:
validateFile()
{
echo "$1" | grep '.zip$' > /dev/null
if [ $? -eq 0 ]
then
echo 1
else
echo 0
fi
}
printf "\n Enter Source File1 Path: "
read file1
res=$(validateFile $file1);
echo "$res"
What you want is maybe
validateFile()
{
echo "$1" | grep '.zip$' > /dev/null
if [ $? -eq 0 ]
then
# return appropriate value
return 1
else
return 0
fi
}
printf "\n Enter Source File1 Path: "
read file1
# call function
validateFile $file1
# use return code
res=$?
echo "$res"
return should be reserved for the exit code of the function and echo should be used to return values.
$( ) construction catching the echo output. If non-recursive there still is the possibility to use a variable to return a value solving all problems at once.$(...) invokes a subshell and captures its standard output. So you can do one of these two things:
foo() {
fooresult=1
}
foo
echo $fooresult
or this:
foo() {
echo 1
}
fooresult=$(foo)
echo $fooresult
validateFile()
{
echo "$1" | grep '.zip$' > /dev/null
if [ $? -eq 0 ]
then
echo 1
else
echo 0
fi
}
printf "\n Enter Source File1 Path: "
read file1
res=$(validateFile $file1);
echo "$res"
