递归查找与特定模式匹配的所有文件

Question

I need to find (or more specifically, count) all files that match this pattern:

*/foo/*.doc

Where the first wildcard asterisk includes a variable number of subdirectories.

Answer 1

With gnu find you can use regex, which (unlike -name ) match the entire path:

find . -regex '.*/foo/[^/]*.doc'

To just count the number of files:

find . -regex '.*/foo/[^/]*.doc' -printf '%i\n' | wc -l

(The %i format code causes find to print the inode number instead of the filename; unlike the filename, the inode number is guaranteed to not have characters like a newline, so counting is more reliable. Thanks to @tripleee for the suggestion.)

I don't know if that will work on OSX, though.

Answer 2

how about:

find BASE_OF_SEARCH/*/foo -name \\*.doc -type f | wc -l

What this is doing:

• start at directory BASE_OF_SEARCH/
• look in all directories that have a directory foo
• look for files named like *.doc
• count the lines of the result (one per file)

The benefit of this method:

• not recursive nor iterative (no loops)
• it's easy to read, and if you include it in a script it's fairly easy to decipher (regex sometimes is not).

UPDATE: you want variable depth? ok:

find BASE_OF_SEARCH -name \\*.doc -type f | grep foo | wc -l

• start at directory BASE_OF_SEARCH
• look for files named like *.doc
• only show the lines of this result that include "foo"
• count the lines of the result (one per file)

Optionally, you could filter out results that have "foo" in the filename, because this will show those too.

Answer 3

Untested, but try:

find . -type d -name foo -print | while read d; do echo "$d/*.doc" ; done | wc -l

find all the "foo" directories (at varying depths) (this ignores symlinks, if that's part of the problem you can add them); use shell globbing to find all the ".doc" files, then count them.

Answer 4

基于此页面上其他页面上的答案，我设法将以下内容放在一起，其中在当前文件夹中执行搜索，在其下的所有其他文件中执行扩展名为pdf的所有文件，然后对包含test_text的文件进行过滤他们的头衔。

find . -name "*.pdf" | grep test_text | wc -l