UNIX Shell腳本

Question

目前我正在研究2代碼：

第一個代碼是向用戶顯示菜單的代碼（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

script.sh的代碼是：

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

基本上，運行script.sh時，我需要輸入1個參數，例如./script.sh home ，它與Place.txt中的“ home”一詞匹配。

但是，當我運行“ start.sh”腳本並選擇運行“ script.sh”腳本的數字1時，我無處可輸入參數（主目錄），“ script.sh”代碼將無法運行。

任何建議如何實現？

Answer 1

您可以從start.sh中讀取輸入：

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

或者在script.sh中執行：

script.sh

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

只是兩者都不做。

Answer 2

為什么要在一個腳本中同時調用另一個shell腳本？

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