CODEWITHSUNDEEP
functionshell
pihug12
pihug12

Reputation: 3

Use a function in another function in shell

Type()
{
    if [ -d $1 ]
        then
            return 1
    elif [ -e $1 ]
        then
            return 2
    else
            return 0
    fi
}

Types()
{
    local arg1 arg2
    for arg1 in $@
        do
            arg2=$(Type $arg1)
            if [ arg2 -eq 1 ]
                then
                    echo "$arg1 est un répertoire."
            elif [ arg2 -eq 2 ]
                then
                    echo "$arg1 n est pas un répertoire."
            else
                    echo "$arg1 ne correspond à aucune entrée du répertoire."
            fi
    done
}

I don't know how can I use the function 'Type' in 'Types'. I tried "arg2=$(Type $arg1)" bur it doesn't seem to work. What's the correct syntax please ?

Upvotes: 0

Views: 62

Answers (2)

PleaseStand
PleaseStand

Reputation: 32082

return 1

Bash functions often return values through their standard output. For example, you could use something like this instead:

exec echo 1    # sends 1 to the standard output and then ends the function

Alternatively, you could return an integer between 0 and 255 as an exit code (as you are trying to do). If you choose to do so, you need to do:

Type $arg1
arg2=$?        # obtains exit code of last command/function executed

However, if you need to return an array, you must use a global variable. You may want to refer to the Complex Functions and Function Complexities section of the Advanced Bash-Scripting Guide for examples of this method.

Upvotes: 1

Jonathan Callen
Jonathan Callen

Reputation: 11571

If you want to use the function with $(Type ...), then you can change the return ... statements to echo ... (not exec echo ..., which does something else). If you want to keep the return ... statements in Type(), then you need to do something like Type ...; arg2=$? to test the return code from Type().

Upvotes: 1

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