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I have 10 files in a dictionary and want to list only files that have an "c" or "z" on position 3 using the ls command. How do I do this?

uni flocks
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2 Answers2

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Simple globbing:

ls -ld -- ??[cz]*

  • ? - matches any single character

  • [cz] - matches one character given in the brackets (character class)

Note that hidden files will not be included.

Stéphane Chazelas
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RomanPerekhrest
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Use globbing, each question mark represents a character. Therefore, the answer below indicates that you want anything listed that contains c or z character as a third occurrence.

ls ??{c*,z*}

or:

ls ??{c,z}*
αғsнιη
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user2062070
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    also note that this require a shell that support brace-expansion, see [Why is brace expansion not supported?](https://unix.stackexchange.com/q/92819/72456) – αғsнιη Oct 29 '21 at 12:49
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    Contrary to Roman's approach, that also means expanding too independent globs so is less efficient and in `zsh` or `bash -O failglob`, that would abort the command if *either* of the two globs has no match. – Stéphane Chazelas Oct 29 '21 at 13:31
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    and in Bash with default settings (`failglob` and `nullglob` not set), this will give an error from `ls` for the nonmatching glob, e.g. if the existing files are `abcde`, `fghij`, `ls` will list `abcde` and give an error for `??z*`. There's really no reason to use braces to single-character alternatives, when brackets already work (like in the other answer). Even if the alternatives did consist of more than one character, ksh-style extended globs would be better (e.g. `??@(c|foo)*`). – ilkkachu Oct 29 '21 at 13:46