修改传递给脚本的参数 (Bash) [英] Modifying a parameter pass to a script (Bash)

问题描述

我已经在 Google 上搜索了很长时间，但找不到与我需要/想做的事情相匹配的任何内容.

I have been looking on Google for quite a while now and can't find anything that is matching what I need/want to do.

我的目标是编写一个接受两个参数的脚本.它将搜索第一个参数(它是一个列表)并检测第二个参数是否已经在其中.例如:

My objective is to write a script that takes two arguments. It will search through the first argument (which is a list) and detect if the second argument is already in it. For example:

list =/bin/foo:/bin/random:random

list = /bin/foo:/bin/random:random

添加到列表:/bin/foobar

to add to list: /bin/foobar

调用脚本会产生/bin/foo:/bin/random:random:/bin/foobar的结果.

Calling the script will produce the result of /bin/foo:/bin/random:random:/bin/foobar.

如果要添加到列表中的部分已经在列表中，则不会对原始内容进行任何更改.

If the part to add to the list is already in the list then nothing will be changed of the original.

直到我想修改我传递的参数为止，我的一切都在工作.

I have everything working up until the point where I want to modify the parameter I passed.

...
if [ $RUN = 1 ]; then
    echo $1
else
    $1="$NEWLIST"
fi
exit 0

然而这产生了错误.它说没有找到该命令，并给了我 $1="$NEWLIST" 所在的行号.我在这里做错了什么?我如何修改 $1?谢谢！

This however produced an error. It says that the command isn't found and gives me the line number that $1="$NEWLIST" is on. What am I doing wrong here? How do I modify $1? Thanks!

$ PATH=/opt/bin:$PATH
$ ./scrip.sh PATH /user/opt/bin
$ /opt/bin:/user/opt/bin

这是我想要的脚本结果.

This is what I would want as a result of the script.

推荐答案

adymitruk 已经说过了，但是为什么要赋值给一个参数.这样做不行吗?

adymitruk already said it, but why do you want to assign to a parameter. Woudln't this do the trick?

if `echo :$1: | grep ":$2:" 1>/dev/null 2>&1`
then
  echo $1
else
  echo $1:$2
fi

也许是这样:

list="1:2:3:4"
list=`./script $list 5`;echo $list

大

使用此脚本(例如称为 listadd):

Use this script (called listadd for instance):

if ! `echo :${!1}: | grep ":$2:" 1>/dev/null 2>&1`
then
  export $1=${!1}:$2
fi

并从您的 shell 中获取它.结果如下(我希望这是 wsa 的意图):

And source it from your shell. Result is the following (I hope this is what wsa intended):

lorenzo@enzo:~$ list=1:2:3:4
lorenzo@enzo:~$ source listadd list 3
lorenzo@enzo:~$ echo $list
1:2:3:4
lorenzo@enzo:~$ source listadd list 5
lorenzo@enzo:~$ echo $list
1:2:3:4:5
lorenzo@enzo:~$ list2=a:b:c
lorenzo@enzo:~$ source listadd list2 a
lorenzo@enzo:~$ echo $list2
a:b:c
lorenzo@enzo:~$ source listadd list2 d
lorenzo@enzo:~$ echo $list2
a:b:c:d

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