bash 在复合命令中查找 + ls -lT
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Find + ls -lT in compound command
提问by David542
To get the full timestamp of a file, I can do:
要获取文件的完整时间戳，我可以执行以下操作：
$ ls -lT
However, when I try the following:
但是，当我尝试以下操作时：
find . -ls -lT
I get an find: -lt: unknown primary or operator(using find . -lsworks).
我得到一个find: -lt: unknown primary or operator（使用find . -ls作品）。
What would be the correct way to use the find+ ls -lTcommand?
使用find+ls -lT命令的正确方法是什么？
回答by Andy Ross
The find "-ls" option isn't running ls and doesn't accept all its arguments. That said, I don't know why you want the -T argument, which is an obscure thing involving tab stops that I had to look up. But broadly, you just want to run a command ("ls -lT" in this case) on a bunch of files found by find. So any of the following should work: find . -type f | xargs -n1 ls -lTor find . -type f -exec ls -lT {} ';'or for i in $(find . -type f); do ls -lT $i; done.
find "-ls" 选项不运行 ls 并且不接受其所有参数。也就是说，我不知道你为什么想要 -T 参数，这是一个涉及制表位的晦涩的东西，我不得不查找。但总的来说，您只想对 find 找到的一堆文件运行一个命令（在本例中为“ls -lT”）。因此，以下任何一项都应该有效：find . -type f | xargs -n1 ls -lTorfind . -type f -exec ls -lT {} ';'或for i in $(find . -type f); do ls -lT $i; done。
Or, for the special case of ls that takes more than one command line argument, just find . -type f | xargs ls -lT
或者，对于需要多个命令行参数的 ls 的特殊情况，只需 find . -type f | xargs ls -lT
回答by Andy Lester
If you're parsing the output of ls, you're doing it wrong. There's always a better way to interface with the filesystem than parsing human-readable output.
如果您正在解析 的输出ls，那么您就做错了。总有比解析人类可读输出更好的方式与文件系统交互。
I'm not sure what you mean by the "full timestamp of the file", but if you want, say, the last modification time, use stat.
我不确定你所说的“文件的完整时间戳”是什么意思，但如果你想要，比如上次修改时间，请使用stat.
stat --printf='%x' foo
2013-01-29 13:33:32.000000000 -0600
Run man statto see all the other formatting options in the --printfargument.
运行man stat以查看参数中的所有其他格式选项--printf。
