bash - 如何在 bash 中打印用法和缺少参数错误？

Question

我有一个需要 5 个参数的 bash 脚本。如果其中任何一个丢失并退出那里，我想打印用法和丢失的参数错误。还有帮助选项，它只打印 $usage。

#script.sh
usage="$0 param1 param2 param3 param4 param5
         param1 is ..
         param2 is ..
         param3 is ..
         param4 is ..
         param5 is .."

#if script.sh
# prints $usage and all param missing

#if script.sh param1 param2 param3
# print $usage and param4 and param5 missing and exit and so on

# script.sh -h
# just print $usage

Answer 1

你可以使用

var=${1:?error message}

它在 $var 中存储 $1 的值(如果已设置)，如果未设置，则显示错误消息并停止执行.

例如:

src_dir="${1:?Missing source dir}"
dst_dir="${2:?Missing destination dir}"
src_file="${3:?Missing source file}"
dst_file="${4:?Missing destination file}"

# if execution reach this, nothing is missing

src_path="$src_dir/$src_file"
dst_path="$dst_dir/$dst_file"
echo "Moving $src_path to $dst_path"
mv "$src_path" "$dst_path"