我如何使用一个变量参数号在bash脚本？ [英] How do I use a variable argument number in a bash script?

问题描述

#!/bin/bash
# Script to output the total size of requested filetype recursively

# Error out if no file types were provided
if [ $# -lt 1 ]
then 
  echo "Syntax Error, Please provide at least one type, ex: sizeofTypes {filetype1} {filetype2}"
  exit 0
fi

#set first filetype
types="-name *."$1

#loop through additional filetypes and append
num=1
while [ $num -lt $# ]
do
  (( num++ ))
  types=$types' -o -name *.'$$num
done

echo "TYPES="$types

find . -name '*.'$1 | xargs du -ch *.$1 | grep total

我有这个问题就在这里：

The problem I'm having is right here:

 #loop through additional filetypes and append
    num=1
    while [ $num -lt $# ]
    do
      (( num++ ))
      types=$types' -o -name *.'>>$$num<<
    done

我只是想通过遍历所有的参数不包括第一位的，应该很容易，但我有困难的时候搞清楚如何使这项工作。

I simply want to iterate over all the arguments not including the first one, should be easy enough, but I'm having a difficult time figuring out how to make this work

推荐答案

从bash的手册页：

  shift [n]
          The  positional  parameters  from n+1 ... are renamed to $1 ....
          Parameters represented by the numbers  $#  down  to  $#-n+1  are
          unset.   n  must  be a non-negative number less than or equal to
          $#.  If n is 0, no parameters are changed.  If n is  not  given,
          it  is assumed to be 1.  If n is greater than $#, the positional
          parameters are not changed.  The return status is  greater  than
          zero if n is greater than $# or less than zero; otherwise 0.

所以，你的循环会是这个样子：

So your loop is going to look something like this:

#loop through additional filetypes and append
while [ $# -gt 0 ]
do
  types=$types' -o -name *.'$1
  shift
done

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