我正在构建一个最小的标准bash菜单。选择工作正常,但命令未执行。我怀疑我可能会包装命令,但是不确定如何完成。我经常会收到2条命令,因此需要以bash将其理解为命令行的方式来分隔命令。
我发现了这样的标题类似的SO问题,但没有回答为什么不执行bash脚本中的命令的原因:
Cannot execute shell commands in bash script
什么是行不通的?
按1:不会更改为正确的cd。
按2:在正确的文件夹中创建文件。
按3:工作。
按4:Works(R文件准备为:a <-1)。
想要的行为:
我需要执行bash菜单脚本中的命令。
#!/bin/bash
# -----
# Menu:
# -----
while :
do
echo "Menu"
echo "1 - Change directory to /tmp"
echo "2 - Create file test1.sh with hello message, in /tmp"
echo "3 - Execute test1.sh"
echo "4 - Execute R-script [/tmp/test2.R]"
echo "Exit - any kind but not [1-4]"
read answer;
case $answer in
1)
echo "Change directory to [\tmp]"
cd /tmp # Command to be executed.
break
;;
2)
echo "Create file [test1.sh] in [\tmp]"
# Commands to be executed.
cd /tmp
touch test1.sh
chmod +x test1.sh
echo echo "hello" > test1.sh
break
;;
3)
echo "Execute file [test1.sh]"
/tmp/./test1.sh # Command to be executed.
break
;;
4)
echo "Execute R-script [/tmp/test2.R]"
cd /tmp && /usr/bin/Rscript test2.R # Command to be executed.
break
;;
*)
# Command goes here
echo "Exit"
break
;;
esac
done
答案 0 :(得分:0)
要执行所有情况时,请勿使用break。
并且在您使用/tmp/./test1.sh
的情况3中,它应该是./tmp/test1.sh
或sh /tmp/test1.sh
您的代码应为:
#!/bin/bash
# -----
# Menu:
# -----
while :
do
echo "Menu"
echo "1 - Change directory to /tmp"
echo "2 - Create file test1.sh in /tmp"
echo "3 - Execute test1.sh"
echo "4 - Execute R-script [/tmp/test2.R]"
echo "Exit - any kind but not [1-4]"
read answer;
case $answer in
1)
echo "Change directory to [\tmp]"
cd /tmp # Command to be executed.
pwd
;;
2)
echo "Create file [test1.sh] in [\tmp]"
touch /tmp/test1.sh # Command to be executed.
;;
3)
echo "Execute file [test1.sh]"
./tmp/test1.sh # Command to be executed.
;;
4)
echo "Execute R-script [/tmp/test2.R]"
/usr/bin/Rscript /production/20_front_trader/build/x_run_front_trader.R # Command to be executed.
;;
*)
# Command goes here
echo "Exit"
;;
esac
done