如何使脚本采用多个参数？

这是一个非常简单的脚本

#!/usr/local/bin/bash
set -e
if [ "$#" -lt 1 ]
then
echo  "Please insert at least one argument"
exit
else
echo -e "\c"
fi


if [ -h  "$1" ]
then
         echo "$1 is a symbolic link"
else    
         echo "$1 in not a symbolic link"
fi

〜
我的问题是：如何更改脚本以识别多个参数？我有4个文件，我希望脚本返回

$1 is a symbolic link
$2 is not a symbolic link
$3 is not a symbolic link

等等

我该怎么办？

使用与原始脚本相同的结构，您只需要遍历$@数组（这是命令行中提供的参数列表）：

#!/usr/local/bin/bash
set -e
if [ "$#" -lt 1 ]
then
echo  "Please insert at least one argument"
exit
else
echo -e "\c"
fi


for file in "$@"
do
    if [ -h  "$file" ]
    then
         echo "$file is a symbolic link"
    else    
         echo "$file is not a symbolic link"
    fi
done

同一件事的简化版本是：

#!/usr/bin/env bash
[ "$#" -lt 1 ] && printf "Please give at least one argument\n" && exit 
for file 
do
    [ -h "$file" ] && printf "%s is a symbolic link\n" "$file" || 
        printf "%s is not a symbolic link\n" "$file"
done

没人提到班次吗？

if [ x = "x$1" ] ; then
    echo need at least one file
    exit 1
fi

while [ x != "x$1" ] ; do
  if [ -h  "$1" ]; then
    echo "$1 is a symbolic link"
  else    
    echo "$1 is not a symbolic link"
  fi
  shift
done

您可以使用for循环来处理传递给脚本的所有文件：

for f do
  if [ -h  "$f" ]; then
    printf "%s is a symbolic link\n" "$f"
  else    
    printf "%s is not a symbolic link\n" "$f"
  fi
done

另一个shift：

: "${1?USAGE: "$0" files...}"
while   [ "$#" -gt 0 ]
do      [ -h "$1" ]
        printf "'%s' is %.$((!$?*4))s%s\n" \
               "$1" "not " "a symbolic link." 
shift;  done