如何在 bash 脚本中使用可变参数编号？
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How do I use a variable argument number in a bash script?
提问by Corbin Tarrant
#!/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 *."
#loop through additional filetypes and append
num=1
while [ $num -lt $# ]
do
(( num++ ))
types=$types' -o -name *.'$$num
done
echo "TYPES="$types
find . -name '*.' | xargs du -ch *. | grep total
The problem I'm having is right here:
我遇到的问题就在这里：
#loop through additional filetypes and append
num=1
while [ $num -lt $# ]
do
(( num++ ))
types=$types' -o -name *.'>>$$num<<
done
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
我只是想迭代所有参数，不包括第一个，应该很容易，但我很难弄清楚如何使这项工作
回答by Zac Thompson
from the bash man page:
从 bash 手册页：
shift [n]
The positional parameters from n+1 ... are renamed to ....
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.
So your loop is going to look something like this:
所以你的循环看起来像这样：
#loop through additional filetypes and append
while [ $# -gt 0 ]
do
types=$types' -o -name *.'
shift
done
回答by amertune
If all you're trying to do is loop over the arguments, try something like this:
如果您要做的只是遍历参数，请尝试以下操作：
for type in "$@"; do
types="$types -o -name *.$type"
done
To get your code working though, try this:
为了让您的代码正常工作，请尝试以下操作：
#loop through additional filetypes and append
num=1
while [ $num -le $# ]
do
(( num++ ))
types=$types' -o -name *.'${!num}
done
回答by ghostdog74
if you don't want to include the first one, the way to do that is to use shift. Or you can try this. imagine variable sis your arguments passed in.
如果您不想包含第一个，那么这样做的方法是使用 shift。或者你可以试试这个。想象变量s是你传入的参数。
$ s="one two three"
$ echo ${s#* }
two three
Of course, this assume you won't be passing in strings that is one word by itself.
当然，这假设您不会传入一个单独的单词的字符串。
