如何使脚本接受多个参数?

如何使脚本接受多个参数?

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

ETC

我怎么能这样做呢?

答案1

使用与原始脚本相同的结构,您只需要迭代数组$@(即命令行中给出的参数列表):

#!/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

答案2

没人提到换班吗?

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

答案3

您可以使用for循环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

答案4

我阅读它的方式是如何让我的个人脚本采用多个参数,例如 wget 可以执行“wget url1 url2 url3”

将整个脚本粘贴到以下内容中,其中显示“来自我的脚本的行”,并将此文件保存为新脚本:

ARGUMENTS=$(echo "$@"| tr " " "\n") ; while read A; do "lines from my script";done < "$ARGUMENTS"

脚本的参数存储在变量 $@ 中。这个小脚本环绕您的脚本并多次运行它,每次将下一个参数传递给它。如果您当前的脚本引用 $1,您必须将其更改为 $A。

相关内容