在目录中找到最新的构建文件

Question

I need to get the latest build file in current directory . The logic is sth like this:

1. Search for pattern in given build directory
2. find the latest one among matched
3. return basename for latest file name

I got this one so far but its not complete

   find  ./build  -iregex '.*/build_.*\.tar\.gz' -type f -exec basename {} \; 

I am puzzled around sorting it to get latest one

Answer 1

This find should work using stat and sort :

find ./build  -iregex '.*/build_.*\.tar\.gz' -type f -exec stat -c '%Y %n' {} + |
    sort -rn -k1,1 | head -1 | cut -d " " -f2-

---

On OSX try this sed :

find ./build  -iregex '.*/build_.*\.tar\.gz' -type f -exec stat -f '%m %N' {} + |
    sort -rn -k1,1 | head -1 | cut -d " " -f2-

Answer 2

For a pure Bash possibility:

#!/bin/bash

shopt -s globstar nullglob nocaseglob

latest=
for file in ./build/**/build_*.tar.gz; do
    [[ -f $file ]] || continue
    [[ $latest ]] || latest=$file
    [[ $file -nt $latest ]] && latest=$file
done

if [[ $latest ]]; then
    echo "Latest build: ${latest##*/}"
else
    echo "No builds found"
fi

Answer 3

Having GNU find , you can use the following command:

find ./build  -iregex '.*/build_.*\.tar\.gz' -printf '%T@ %f\n' | sort -n | tail -n1 | cut -d' ' -f2

It uses find 's printf action to print the timestamps of the latest modification along with the filename (basename). Then it pipes it to sort , extracts the last line using tail and finally separates the name from the timestamp using cut .