bash Unix 查找：来自标准输入的文件列表

Unix find: list of files from stdin

提问by JXG

I'm working in Linux & bash (or Cygwin & bash).

我在 Linux 和 bash（或 Cygwin 和 bash）中工作。

I have a huge--huge--directory structure, and I have to find a few needles in the haystack.

我有一个巨大的——巨大的——目录结构，我必须在大海捞针中找到几根针。

Specifically, I'm looking for these files (20 or so):

具体来说，我正在寻找这些文件（20 个左右）：

foo.c
bar.h
...
quux.txt

I know that they are in a subdirectory somewhere under ..

我知道它们位于..

I know I can find any one of them with find . -name foo.c -print. This command takes a few minutes to execute.

我知道我可以找到其中的任何一个 find . -name foo.c -print。此命令需要几分钟才能执行。

How can I print the names of these files with their full directory name? I don't want to execute 20 separate finds--it will take too long.

如何使用完整目录名打印这些文件的名称？我不想执行 20 个单独find的 s——它会花费太长时间。

Can I give findthe list of files from stdin? From a file? Is there a different command that does what I want?

我可以提供find来自标准输入的文件列表吗？从文件？是否有不同的命令可以执行我想要的操作？

Do I have to first assemble a command line for findwith -ousing a loop or something?

我必须先组装一个命令行对find与-o使用循环或东西吗？

采纳答案by jm666

If your directory structure is huge but not changing frequently, it is good to run

如果你的目录结构很大但不经常变化，运行就好

cd /to/root/of/the/files
find . -type f -print > ../LIST_OF_FILES.txt #and sometimes handy the next one too
find . -type d -print > ../LIST_OF_DIRS.txt

after it you can really FAST find anything (with grep, sed, etc..) and update the file-lists only when the tree is changed. (it is a simplified replacement if you don't have locate)

在它之后，您可以真正快速地找到任何东西（使用 grep、sed 等）并仅在树更改时更新文件列表。（如果你没有，它是一个简化的替代品locate）

So,

所以，

grep '/foo.c$' LIST_OF_FILES.txt #list all foo.c in the tree..

When want find a list of files, you can try the following:

当想要查找文件列表时，您可以尝试以下操作：

fgrep -f wanted_file_list.txt < LIST_OF_FILES.txt

or directly with the find command

或直接使用 find 命令

find . type f -print | fgrep -f wanted_file_list.txt

the -ffor fgrep mean - read patterns from the file, so you can easily grepping input for multiple patterns...

在-f对fgrep一样均值-从文件中读取模式，让您可以轻松grepping输入的多模式...

回答by pavium

You shouldn't need to run findtwenty times.

你不应该需要跑find二十次。

You can construct a single command with a multiple of filename specifiers:

您可以使用多个文件名说明符构造单个命令：

find . \( -name 'file1' -o -name 'file2' -o -name 'file3' \) -exec echo {} \;

回答by sarnold

Is the locate(1)command an acceptable answer? Nightly it builds an index, and you can query the index quite quickly:

是locate(1)命令一个可以接受的答案？每晚它都会构建一个索引，您可以非常快速地查询索引：

$ time locate id_rsa
/home/sarnold/.ssh/id_rsa
/home/sarnold/.ssh/id_rsa.pub

real    0m0.779s
user    0m0.760s
sys 0m0.010s

I gave up executing a similar findcommand in my home directory at 36 seconds. :)

我find在 36 秒时放弃了在我的主目录中执行类似的命令。:)

If nightly doesn't work, you could run the updatedb(8)program by hand once before running locate(1)queries. /etc/updatedb.conf(updatedb.conf(5)) lets you select specific directories or filesystem types to include or exclude.

如果 nightly 不起作用，您可以updatedb(8)在运行locate(1)查询之前手动运行该程序一次。/etc/updatedb.conf( updatedb.conf(5)) 允许您选择要包含或排除的特定目录或文件系统类型。

回答by Ignacio Vazquez-Abrams

Yes, assemble your command line.

是的，组装你的命令行。

回答by marco

Here's a way to process a list of files from stdin and assemble your (FreeBSD) find command to use extended regular expression matching (n1|n2|n3).

这是一种处理来自 stdin 的文件列表并组合您的 (FreeBSD) find 命令以使用扩展正则表达式匹配的方法(n1|n2|n3)。

For GNU find you may have to use one of the following options to enable extended regular expression matching:

对于 GNU find，您可能必须使用以下选项之一来启用扩展正则表达式匹配：

-regextype posix-egrep

-regextype posix-egrep

-regextype posix-extended

-regextype posix-extended

echo '
foo\.c
bar\.h
quux\.txt
' | xargs bash -c '
IFS="|"; 
find -E "$PWD" -type f -regex "^.*/($*)$" -print
echo find -E "$PWD" -type f -regex "^.*/($*)$" -print
' arg0

# note: "$*" uses the first character of the IFS variable as array item delimiter
(
IFS='|'
set -- 1 2 3 4 5
echo "$*"   # 1|2|3|4|5
)