if condition implementation in linux shell script

Question

I have a list of 100+ folders in a linux at a folder path "files/data" where I already have basic script "list.sh".

for d in files/data/*
do
echo $d
done

When I run it...

sh list.sh

It is listing all the 100+ folders (amazon, apple, cola, pepsi, and etc...) in side the "files/data" folder. But I am looking for some if command implementation to achieve the bellow.

When I call...

sh list.sh --brand=apple

Then it should only list "apple" and if i call...

sh list.sh --brand=all

Then it should list all the 100+ folders (amazon, apple, cola, pepsi, and etc...).

Thanks.

Community · Accepted Answer · 2020-06-20 09:12:55Z

1

I've edit your's script

Usage:

./script.sh all - show all dirs in folder path

./script.sh apple - show apple dirs

You can't use script without parameter because usage of grep.

Hope that's all what you need

#!/bin/bash

brand=$1


function ShowDirs()
{
for d in files/data/*
do
    if ! [[ $brand == all ]];
     then
        echo $d | grep $brand
    else
        echo $d
    fi
done
}

ShowDirs

edited Jun 20, 2020 at 9:12

CommunityBot

11 silver badge

answered Jan 4, 2018 at 9:55

Kend3r

263 bronze badges

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