如何从两个列表中删除列表中的目录? [英] How to remove directories in a list from two lists?

问题描述

我正在编写一个 c-shell 脚本，其中我在两个字符串中查找两个不同的目录.我想删除相同的目录的名称.我只想要两者中的唯一目录，而忽略重复的目录.我对如何做到这一点感到有些困惑.

I am writing a c-shell script, where I am grepping two different directories in two strings. I wanted to remove the name of the directories which are the same. I only want the unique directory among the two by leaving out the duplicate ones. I am little confused about how to do this.

sta_views 和 pnr_views 字符串中有一些常见的目录.我使用 ls -l 命令对他们两个进行了搜索，如上所示存储了它们.现在我想要做的是首先我们可以遍历 sta_view 并检查 pnr_view 列表中是否存在它，如果不存在，则将它们放在一个单独的列表中，如果是，则不要做任何事情"；希望这有助于您理解我的问题.谢谢！

There are some common directories present in sta_views and pnr_views strings. I am grepped both of them using ls -l command as seen above stored them. Now what I want to do is "first we can loop over sta_view and check if that exist in the pnr_view list and if no , then put them in a separate list , if yes , then do not do anything" Hope this helps to understand you my question. Thank You!

请告诉我这样做的方法，

Please let me know the approach to do it,

set pnr_views = `(ls -l pnr/rep/signoff_pnr/ | grep '^d' | awk '{print $9}'
)`
set sta_views = `(ls -l sta/rep/ | grep '^d' | grep -v common | grep -v signoff.all.SAVEDESIGN | awk '{print $9}'
)`

foreach i ($sta_view) 
if [$i == $pnr_view] #just want to remove the list pnr_view from sta_view
then
echo ...
else 
 

此处sta_views 包含存在于pnr_views 中的目录.如何从 sta_views 中删除 pnr_views 目录?

Here sta_views contains the directories that are present in pnr_views. How shall I remove the pnr_views directories from sta_views?

推荐答案

循环第一个目录中的目录，只需检查第二个目录中是否存在相同的条目.

Loop over the directories in the first directory, and simply check whether the same entry exists in the second.

foreach i ( pnr/rep/signoff_pnr/*/ )
  if ( -d sta/rep/"$i:h:t" ) then
    switch ("$i:h:t")
      case *'common'*:
      case *'signoff.all.SAVEDESIGN'*:
        breaksw
      default:
        echo sta/rep/"$i:h:t"
        breaksw
    endsw
  endif
end

Csh 变量修饰符$i:h 返回$i 中文件名的目录名部分，修饰符:t 返回最后一个元素.

The Csh variable modifier $i:h returns the directory name part of the file name in $i and the modifier :t returns the last element from that.

Csh 非常善变，因此现在相当不受欢迎；如果您切换到与 Bourne 兼容的现代 shell，您的运气可能会更好，特别是如果您只是在学习.另请参阅 Csh 被认为有害 以及 https://www.shlomifish.org/open-source/anti/csh/.循环 bash 中的值对 展示了如何在重击.

Csh is extremely fickle and thus these days rather unpopular; you might have better luck if you switch to a modern Bourne-compatible shell, especially if you are only just learning. See also Csh considered harmful and the other links on https://www.shlomifish.org/open-source/anti/csh/. Looping over pairs of values in bash shows how to do something similar in Bash.

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