Question

我想列出从现在的目录开始的所有目录。我不想把先前的名录(.)列入,因此我有~~的常规表述~~。我正在使用sh子

ls .[^.]*

该公司在UNIX变量平台上进行罚款,但在太阳系中,我仍在使用这个薄子,但我只有目录。我怎么能具体说明太阳系的确切功能 ~~正式表述~~。它用巴什开展工作,但我只需要使用sh子,因为保证能够在我们org的每个安全研究所盒子上找到。

Answer 1

You can do this with /bin/sh, but with a slightly different syntax; it uses ! rather than ^ to negate the set of characters:

ls .[!.]*

Note that this is a file matching pattern (a glob), not a regular expression.

但是,鉴于你在由各种炮弹支持的glob syntax中再次看到的变化,你可能最好使用<条码>grep,并使用实际的定期表述。

Answer 2

www.un.org/Depts/DGACM/index_spanish.htm ow,我不知道他们仍然说: 页: 1 因此,我不知道太阳星号()。 cr:

ls `ls -a | grep  ^.[^.].* `

The -a for ls tells it to include all the dot-files in the list and then a bit of grep for dealing with the dots. That matches what ls .[^.]* does in OSX s csh, hopefully it will work with Solaris s as well.

Answer 3

Character classes in globs in csh don t support negation...

• 从太阳的人类网页上:

 [...] Match any single character in the enclosed list(s)  or
       range(s).  A  list is a string of characters.  A range
       is  two  characters  separated  by  a  dash  (-),  and
       includes  all  the  characters in between in the ASCII
       collating sequence (see ascii(5)).

因此,你可以创建一种特性类别,包括:40(空间)到(......之前的特性,从/到(其余可打印的正文)。

ls -a .[ --/-~]*

This seems to work for me, but I haven t checked the output with a fine-toothed comb.

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