Linux 如何递归列出具有大小和上次修改时间的文件？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/6964565/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to recursive list files with size and last modified time?
提问by whatupdave
Given a directory i'm looking for a bash one-liner to get a recursive list of all files with their size and modified time tab separated for easy parsing. Something like:
给定一个目录，我正在寻找一个 bash one-liner 来获取所有文件的递归列表，它们的大小和修改的时间选项卡分开以便于解析。就像是：
cows/betsy 145700 2011-03-02 08:27
horses/silver 109895 2011-06-04 17:43
采纳答案by Adam Rosenfield
You can use stat(1)to get the information you want, if you don't want the full ls -loutput, and you can use find(1)to get a recursive directory listing. Combining them into one line, you could do this:
stat(1)如果您不想要完整的ls -l输出，您可以使用来获取所需的信息，并且可以使用find(1)来获取递归目录列表。将它们组合成一行，你可以这样做：
# Find all regular files under the current directory and print out their
# filenames, sizes, and last modified times
find . -type f -exec stat -f '%N %z %Sm' '{}' +
If you want to make the output more parseable, you can use %minstead of %Smto get the last modified time as a time_tinstead of as a human-readable date.
如果您想让输出更易于解析，您可以使用%m而不是%Sm将上次修改时间作为time_t人类可读的日期而不是获取。
回答by John Kugelman
find is perfect for recursively searching through directories. The -lsaction tells it to output its results in ls -lformat:
find 非常适合递归搜索目录。该-ls操作告诉它以以下ls -l格式输出结果：
find /dir/ -ls
On Linux machines you can print customized output using the -printfaction:
在 Linux 机器上，您可以使用以下-printf操作打印自定义输出：
find /dir/ -printf '%p\t%s\t%t\n'
See man findfor full details on the format specifiers available with -printf. (This is not POSIX-compatible and may not be available on other UNIX flavors.)
有关man find可用的格式说明符的完整详细信息，请参阅-printf。（这与 POSIX 不兼容，并且在其他 UNIX 版本上可能不可用。）
回答by Keith Thompson
find * -type f -printf '%p\t%s\t%TY-%Tm-%Td %Tk:%TM\n'
find * -type f -printf '%p\t%s\t%TY-%Tm-%Td %Tk:%TM\n'
If you prefer fixed-width fields rather than tabs, you can do things like changing %sto %10s.
如果您更喜欢固定宽度的字段而不是选项卡，您可以执行诸如更改%s为%10s.
I used find * ...to avoid the leading "./" on each file name. If you don't mind that, use .rather than *(which also shows files whose names start with .). You can also pipe the output through sed 's/^\.\///'.
我曾经find * ...避免在每个文件名中使用前导“./”。如果您不介意，请使用.而不是*（它还显示名称以 开头的文件.）。您还可以通过管道输出sed 's/^\.\///'。
Note that the output order will be arbitrary. Pipe through sortif you want an ordered listing.
请注意，输出顺序将是任意的。sort如果您想要有序列表，请通过管道。
回答by bunk3m
You could try this for recursive listing from current folder called "/from_dir"
您可以尝试从名为“/from_dir”的当前文件夹中递归列表
find /from_dir/* -print0 | xargs -0 stat -c “%n|%A|%a|%U|%G” > permissions_list.txt?
?Lists files and directories passes through to stat command and puts all the info into a file called permissions_list.txt?
? 列出文件和目录传递给 stat 命令并将所有信息放入一个名为 permissions_list.txt 的文件中？
“%n|%A|%a|%U|%G” will give you the following result in the file:
“%n|%A|%a|%U|%G”将在文件中给出以下结果：
from_? dir|drwxr-sr-x|2755|root|root?
from_dir/filename|-rw-r–r–|644|root|root
?Cheers!?
？干杯！？
