Answers:
您可以使用find获取文件列表并执行zip -j myfiles打包操作而忽略路径:
find . -name "d[014]" -exec zip -j myfiles {} +
$ tree
.
├── d0
├── f0
├── f1
│   └── d1
└── f2
    └── f3
        ├── d2
        ├── d3
        └── d4
$ find . -name "d[014]" -exec zip -j myfiles {} +
  adding: d1 (stored 0%)
  adding: d4 (stored 0%)
  adding: d0 (stored 0%)
$ unzip -l myfiles.zip
Archive:  myfiles.zip
  Length      Date    Time    Name
---------  ---------- -----   ----
        0  2017-10-09 10:47   d1
        0  2017-10-09 10:48   d4
        0  2017-10-09 10:47   d0
---------                     -------
        0                     3 files
但是,这仅适用于文件,目录将被忽略zip -j。为了使该功能也适用于目录,请说我们要打包d0,d1而f3在上例中是整个目录,该find行变得稍微复杂一点:
$ find . \( -name "d[01]" -o -name "f3" \) -exec sh -c 'p=$(pwd); for i in $0 $@; do cd ${i%/*}; zip -ur "$p"/myfiles ${i##*/}; cd "$p"; done' {} +
        zip warning: /home/dessert/myfiles.zip not found or empty
  adding: d1 (stored 0%)
  adding: f3/ (stored 0%)
  adding: f3/d3 (stored 0%)
  adding: f3/d2 (stored 0%)
  adding: f3/d4 (stored 0%)
  adding: d0 (stored 0%)
$ unzip -l myfiles.zip 
Archive:  myfiles.zip
  Length      Date    Time    Name
---------  ---------- -----   ----
        0  2017-10-11 10:18   d1
        0  2017-10-11 10:19   f3/
        0  2017-10-11 10:19   f3/d3
        0  2017-10-11 10:19   f3/d2
        0  2017-10-11 10:19   f3/d4
        0  2017-10-11 10:17   d0
---------                     -------
        0                     6 files
zip文件中-j。