如何根据名称模式合并文件

Question

/myfiles/sandboxserver_local.tar.gz.part-aa
/myfiles/sandboxserver_local.tar.gz.part-ab
/myfiles/sandboxserver_local.tar.gz.part-ac

/myfiles/sandboxserver_shared.tar.gz.part-aa
/myfiles/sandboxserver_shared.tar.gz.part-ab
/myfiles/sandboxserver_shared.tar.gz.part-ac

/myfiles/sandboxserver_admin.tar.gz.part-aa
/myfiles/sandboxserver_admin.tar.gz.part-ab
/myfiles/sandboxserver_admin.tar.gz.part-ac

/myfiles/prodserver_local.tar.gz.part-aa
/myfiles/prodserver_local.tar.gz.part-ab
/myfiles/prodserver_local.tar.gz.part-ac

/myfiles/prodserver_shared.tar.gz.part-aa
/myfiles/prodserver_shared.tar.gz.part-ab
/myfiles/prodserver_shared.tar.gz.part-ac

/myfiles/prodserver_admin.tar.gz.part-aa
/myfiles/prodserver_admin.tar.gz.part-ab
/myfiles/prodserver_admin.tar.gz.part-ac

所有这些文件都存储在一个目录中。我想根据其名称模式合并这些文件，如 sandboxserver_local.tar.gz、sandboxserver_shared.tar.gz、sandboxserver_admin.tar.gz、prodserver_local.tar.gz、prodserver_shared.tar.gz、prodserver_admin.tar.gz... ..........有人可以帮忙......提前致谢

Answer 1

#!/usr/bin/env bash

# loop over all files
for file in /myfiles/*.tar.gz.part*; do
   # remove last characters (".part-xx")
   # this creates filename.tar.gz
   tarball="${files%.part-[a-z][a-z]}"
   # if the tarball exists, move to the next
   [[ -f "${tarball}" ]] && continue
   # concatenate tar files
   cat ${tarball}.part-[a-z][a-z] > "${tarball}"
done

Answer 2

find ./myfiles -type f -regex '.*/.*\.tar\.gz\.part-.*' |
sed 's/\.part-.*$//' | 
sort -u |
xargs -n1 -d'\n' -- \
sh -c 'cat "".part-* > ""' --

从名为 *.tar.gz.part-* 的 myfiles 文件夹中获取所有文件。我从文件名中删除最后一个 .part-* 并从列表中仅获取唯一的文件名。然后对于每一行我运行 cat $line.part-* > $line 使用 xargs.

在线版本可在 tutorialspoint 获得。

如何根据名称模式合并文件

how to merge files based on the name pattern

unix

bash

shell

cat