UNIX Shell脚本

Question

currently I'm working on 2 code :

the first code is a code to display a menu to the user (start.sh) :

#!/bin/sh
echo choose a number :
echo "1.)display place information"
echo "2.)look for first character entered"

read number

case $number in
"1") ./script.sh;;
"2") ./script2.sh;;
esac
exit

the code for script.sh is :

#!/bin/sh
grep "$1" Place.txt

basically, when running script.sh , I need to enter 1 argument like ./script.sh home and it match the word 'home' in Place.txt.

But , when i run the "start.sh" script and choose number 1 which run "script.sh" script I have nowhere to enter the argument (home) and the "script.sh" code won't run.

any suggestion how to achieve that?

Answer 1

You can either read the input from start.sh:

start.sh:

#!/bin/sh
echo choose a number :
echo "1.)display place information"
echo "2.)look for first character entered"

read number

case $number in
"1") echo "Please input argument"; read p; ./script.sh "$p";;
"2") echo "Please input argument"; read p; ./script2.sh "$p";;
esac
exit

Or do it in script.sh:

script.sh

#!/bin/sh
echo "Please input argument"
read p
grep "$p" Place.txt

Just dont do both.

Answer 2

why do you want to call another shell script when you can do it in one?

echo choose a number :
echo "1.)display place information"
echo "2.)look for first character entered"

read number

case $number in
1) read -p "Enter pattern to search: " search
   grep $search file.txt
   ;;
2) ;;
esac
exit