awk '{打印$9}'最后一个ls -l列包含文件名中的任何空格

How would I get awk to output $9 $10 $11 etc as some of my files have spaces in them.

我如何得到awk输出$9 $10 $11等等，因为我的一些文件中有空格。

ls -l | grep ^- | awk '{print $9}'

7 个解决方案

#1

A better solution: Don't attempt to parse ls output in the first place.

一个更好的解决方案:首先不要尝试解析ls输出。

The official wiki of the irc.freenode.org #bash channel has an explanation of why this is a Bad Idea, and what alternate approaches you can take instead: http://mywiki.wooledge.org/ParsingLs

irc.freenode.org #bash频道的官方wiki解释了为什么这是一个坏主意，以及可以采用什么替代方法:http://mywiki.wooledge.org/ParsingLs

Use of find, stat and similar tools will provide the functionality you're looking for without the pitfalls (not all of which are obvious -- some occur only when moving to platforms with different ls implementations).

使用find、stat和类似工具将提供您正在寻找的功能，而不存在缺陷(并非所有缺陷都很明显——有些缺陷只在迁移到具有不同ls实现的平台时出现)。

For your specific example, I'm guessing that you're trying to find only files (and not directories) in your current directory; your current implementation using ls -l is buggy, as it excludes files which have +t or setuid permissions. The Right Way to implement this would be the following:

对于您的特定示例，我猜测您试图在当前目录中只查找文件(而不是目录);使用ls -l的当前实现存在bug，因为它排除了具有+t或setuid权限的文件。实现这一目标的正确方法如下:

find . -maxdepth 1 -type f -printf '%f\n'

#2

Just for completion. It can also be done with sed:

只是为了完成。还可以用sed完成:

# just an exercise in regex matching ...
ls -l | sed -E  -e 1d -e /^[^-]/d -e 's/^([^ ]+ +){8}//'

#3

There's probably a better approach that involves combining fields somehow, but:

也许有一种更好的方法可以结合不同的领域，但是:

$ echo 1 2 3 4 5 6 7 8 9 10 11 12 13 14... | awk '{for (i = 9 ; i <= NF ; i++) printf "%s ", $i}'
9 10 11 12 13 14...

Using printf "%s " $i will print the i-th field with a space after it, instead of a newline. The for loop just says to go from field 9 to the last field.

使用printf“%s”$，我将在第i个字段后面加上空格，而不是换行。for循环说从第9场到最后一个场。

#4

echo 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 | awk 'BEGIN {OFS=ORS=""} ; {for (i=9;i<NF;i++) print $i " "; print $NF "\n"}'

#5

If you still insist on the ls -l instead of find or other tools, here is my solution. It is not pretty and destructive:

如果您仍然坚持使用ls -l而不是find或其他工具，这是我的解决方案。它并不美丽和具有破坏性:

Destroy $1 .. $8 by setting them to "" via a for loop
摧毁美元1 . .$8，通过for循环将它们设置为“”
That leaves a bunch of spaces preceding $9, remove them using the sub() command
这就在$9之前留下了一串空格，使用sub()命令删除它们
Print out the remaining

打印出剩下的

ls -l | awk '{for (i = 1; i < 9; i++) $i = ""; sub(/^ */, ""); print}'

l | awk '{for (i = 1;我< 9;我+ +)$ = " ";子(/ ^ * / " ");打印}’

#6

A solution is to encode & decode the space with a word or character by using sed:

一种解决方案是使用sed对空格进行编码和解码:

ls -l | grep ^- | sed s/\ /{space}/ | awk '{print $9}' | sed s/{space}/\ /

This will replace all spaces in a line with {space} before passing it to awk. After the line has passed to awk, we replace {space} back with space.

这将用{space}替换行中的所有空格，然后将其传递给awk。行传递给awk后，我们用空格替换{space}返回。

find as stated by others is a much better solution. But if you really have to use awk, you can try this.

发现别人所说的是一个更好的解决方案。但是如果你真的需要awk，你可以试试这个。

#7

ls -l | grep ^- | awk -v x=9 '{print $x}'

I use this for getting filenames of a directory without incident. I noted the find solution which is fine and dandy if you are unsure of the file types, if you know what you are looking at the ls -l works just fine, it also alphabetically orders by default.

我用它来获取目录的文件名，而不会发生意外。我提到了find解决方案，如果你不确定文件类型，如果你知道你在看什么ls -l很好，它也是默认的字母顺序。

#1