Get only the latest version of a file from across mutiple directories
bash$ find . -name custlist\* | perl -ne '$path = $_; s?.*/??; $name = $_; $map{$name} = $path; ++$c; END { print $map{(sort(keys(%map)))[$c-1]} }'
Explanation
The purpose of this is to find the the “latest” version of the custlist_*.xls
file from among multiple versions in directories and sub-directories, for example:
./c/custlist_v1.003.xls
./c/custlist_v2.001.xls
./d/b/custlist_v1.001.xls
./d/custlist_v1.002.xls
Let’s decompose the one-liner to the big steps:
find . -name custlist\*
— find the files matching the target pattern... | perl -ne '...'
— run perl, with the input wrapped around in awhile
loop so that each line in the input is set in the variable$_
$path = $_; s?.*/??; $name = $_;
— save the full path in$path
, and cut off the subdirectory part to get to the base name of the file and save it in$name
$map{$name} = $path;
— build a mapping of$name
to$path
++$c;
— we count the elements, to use it later(sort(keys(%map)))[$c-1]
— sort the keys of the map, and get the last element, which iscustlist_v2.001.xls
in this exampleEND { print $map{$last} }'
— at the end of all input data, print the path of the latest version of the file
Limitations
Even if the latest version of the file appears multiple times in the directories, the one-liner will print only one of the paths. This could be fixed though if needed.
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