Can I select only one result from a bash glob?

Question

I'm trying to write a script for work to automate some reporting on an output. The Log files are (currently, it's being 'standardise' in the future) stored in this sort of path structure:

/<root_path>/<process_one_path>/logs/<time_date_stamp>/<specific_log_file>

/<root_path>/<process_two_path>/logs/<different_time_date_stamp>/<specific_log_file>

Every part of the path is known except the time date stamps, which are always the latest in the folder.

If I try to use a wild card in place of the time date stamp, I get multiple results, e.g.:

> ls /<root_path>/<process_two_path>/logs/* [tab]
20130102-175103
20130118-090859
20130305-213506

I only want it to return the latest one, is this possible with Bash?

NB (I don't have zsh, and as lovely as it sounds I doubt we'll ever get it at work)

score 11 · Accepted Answer · edited Apr 05 '13 at 22:26

11

The following works in bash 4.2:

list=( /<root_path>/<process_two_path>/logs/* )
echo "${list[-1]}"

If your bash is an older version:

list=( /<root_path>/<process_two_path>/logs/* )
echo "${list[${#list[@]}-1]}"

edited Apr 05 '13 at 22:26

Stéphane Chazelas

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answered Apr 05 '13 at 20:59

phemmer

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This definitely works on my home computer but I can't guarantee it'll work at work? – AncientSwordRage Apr 05 '13 at 21:04
@Pureferret the part that might not work is the negative array index (`-1`). This was added in bash 4.2. RHEL5 is probably the oldest supported mainstream enterprise distro available and it uses bash 4.2.20. – phemmer Apr 05 '13 at 21:14
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Older versions of `bash` have: `"${list[@]: -1}"` like in `ksh93` (`[-1]` comes from `zsh`). – Stéphane Chazelas Apr 05 '13 at 21:26
@Patrick - RHEL5 uses bash `3.2`. RHEL 3 and 4 still supported under extended life. – jordanm Apr 05 '13 at 21:51
@jordanm oh, damn. Well I guess I know where to not look for package versions. Thanks :-) – phemmer Apr 05 '13 at 22:01
@StephaneChazelas That doesn't work when I try it. – phemmer Apr 05 '13 at 22:03
@Patrick, what have you tried. That syntax exists as far back as bash 2.0.3 (15 years ago) at least. Note that you need the space before "-1" to disambiguate from the `${var:-default}` syntax, – Stéphane Chazelas Apr 05 '13 at 22:14
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@Pureferret You can also do arithmetic with the array length, see my edit. – Gilles 'SO- stop being evil' Apr 05 '13 at 22:15
@Gilles, inside `[...]` arithmetic expansion is already on so you don't need the `$((...))`. `${#list}` is correct in `zsh` but not in `ksh93` or `bash` where you need `${#list[@]}` – Stéphane Chazelas Apr 05 '13 at 22:19
@StephaneChazelas Oops, thanks for the reminder, I always get bitten by that one. – Gilles 'SO- stop being evil' Apr 05 '13 at 22:22
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Is it possible to do it inline? That is without intermediate $list variable. – woky Oct 17 '16 at 09:13
@woky I can't think of a way. You could use `echo /foo/* | awk '{ print $1 }'`, but that's pretty dirty, and error prone (spaces break it, etc). You could probably do it in `zsh` though. – phemmer Oct 17 '16 at 12:18

Stéphane Chazelas · Answer 2 · 2022-05-13T08:54:17.567

POSIXly:

 set -- /<root_path>/<process_two_path>/logs/*
 shift "$(($# - 1))"
 printf '%s\n' "$1"

Since you mention zsh:

 print -r /<root_path>/<process_two_path>/logs/*([-1])

If there's no non-hidden file in /<root_path>/<process_two_path>/logs, the POSIX one will output a literal /<root_path>/<process_two_path>/logs/* while zsh will return an error and abort the shell process as if it was a syntax error¹. You can always handle the error in an always block, or use the N (for nullglob) glob qualifier and handle the case by hand yourself.

^{¹ If the glob is in an argument to an external command (which print isn't as it's a zsh builtin), that only exits the process that would have run the command so it just appears as cancelling the command. In interactive shells, it returns to the prompt rather than exiting the shell, same as for other syntax errors.}

Lesmana · Answer 3 · 2022-05-16T21:13:32.890

You can (mis)use a for loop. Like this:

for filename in <logdir>/*; do :; done; echo "$filename"

This works because the loop variable filename still contains the last value after the loop has terminated. Also because bash does glob expansion sorted alphabetically. Read the bash manual for more information. Note that this depends on LC_COLLATE.

This might not be the most efficient because bash still has to loop over all the files just to do nothing. But for reasonable number of files it should still be fast enough. If it is no longer fast enough then the number of files has become unreasonable and you should consider a different approach.

If you want the filename with the oldest timestamp (the one sorted first) you can do it like this:

for filename in <logdir>/*; do echo "$filename"; break; done

or this

for filename in <logdir>/*; do break; done; echo "$filename"

which is effectively the same (the first filename is printed) but depending on your code and style preferences you might prefer one or the other.

score 1 · Answer 4 · answered Apr 06 '13 at 02:44

1

-1 is the default in pipelines, and ls output should already be sorted:

ls | tail -n1

If others were looking for how to insert the first or last result interactively, you can bind menu-complete or menu-complete-backward in .inputrc:

"\e\t": menu-complete
"\e[Z": menu-complete-backward # shift-tab

If show-all-if-ambiguous is enabled, set completion-query-items 0 removes the prompt when there are 101 or more results and set page-completions off disables the pager.

answered Apr 06 '13 at 02:44

Lri

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Parsing `ls` is generally a bad idea -- filenames can contain all sorts of weird characters, including anything `ls` is able to use as a separator. – Gavin S. Yancey Sep 11 '20 at 21:55

score -1 · Answer 5 · edited Apr 06 '13 at 08:07

-1

ls -tr | tail -1

Should do the trick.

edited Apr 06 '13 at 08:07

rahmu

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answered Apr 06 '13 at 06:23

slipset

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Note that you [shouldn't parse the output of `ls`](http://mywiki.wooledge.org/ParsingLs). Ever. – rahmu Apr 06 '13 at 08:09
Why is that? And how does it differ from using the output from a glob? – slipset Apr 06 '13 at 14:37
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http://mywiki.wooledge.org/ParsingLs explains it... – slipset Apr 06 '13 at 14:41

Can I select only one result from a bash glob?

5 Answers5