问题描述
我需要一个 Unix 命令，它将显示我的组列表，用逗号分隔.所以，groups $FT_USER 只显示列表，但我需要把它分成一个逗号分隔的列表！
I need an Unix command which will display a list of my groups separated with commas. So, groups $FT_USER is only displaying the list, but I need to separate it into a comma-separated list!
推荐答案
请注意，这不适用于 Cygwin.实际上，只要名称中有空格的组就行不通.我认为这通常只发生在 Windows (Cygwin) 上，但会导致不正确的结果.我会很快给出一个导致问题的例子.
Note that this will not work for Cygwin. Actually, it won't work whenever there are groups with spaces in the names. I think this usually only happens on Windows (Cygwin), but it leads to incorrect results. I'll quickly give an example where it causes a problem.
$ groups $FT_USER
Users CONSOLE LOGON Authenticated Users CurrentSession LOCAL
这里的组是 {'Users', 'CONSOLE LOGON', 'Authenticated Users', 'CurrentSession', 'LOCAL'}.相信我，或者去看看这篇文章，那里有更多细节.
The groups here are {'Users', 'CONSOLE LOGON', 'Authenticated Users', 'CurrentSession', 'LOCAL'}. Trust me on this one, or go see this post, where there are some more details.
@Kusalananda 之前给出的命令在大多数情况下都有效，但给出了以下不正确的输出.
The command given before by @Kusalananda, which will work in most cases, gives the following, incorrect output.
$ groups $FT_USER | tr ' ' ','
Users,CONSOLE,LOGON,Authenticated,Users,CurrentSession,LOCAL
来自@SriniV 的命令将给出相同的输出.一些组名是分裂的".我们得到 ...,CONSOLE,LOGIN,... 而不是 ...,CONSOLE LOGIN,... 和 ...,Authenticated,Users,... 而不是 ...,Authenticated Users,...
The command from @SriniV will give the same output. Some of the group names are "split". We get ...,CONSOLE,LOGIN,... instead of ...,CONSOLE LOGIN,... and ...,Authenticated,Users,... instead of ...,Authenticated Users,...
(我知道，我知道:你为什么使用 Cygwin?"答案:工作需要它.)
(I know, I know: "Why are you using Cygwin?" Answer: The job requires it.)
要获得我们想要的输出，我们需要使用id 命令，它需要一个 --zero 标志:
To get the output we want, we need to use the id command, which takes a --zero flag:
$ man id | grep -A 1 "\-\-zero"
-z, --zero
delimit entries with NUL characters, not whitespace;
然后我们可以使用@Kusalananda 所描述的 tr 或 awk 和@SriniV，分别.一个问题是，由于尾随NUL，总是会有尾随的,.这可以通过 sed 命令纠正，我们打印除最后一个逗号之外的所有内容，sed 's/\(.*\),/\1/'.请注意， .* 将打印所有内容，直到最后一个逗号.\1 打印(反斜杠转义)括号中的所有内容，即 \1 将打印 .* 部分找到的 >\(.*\),
We can then use tr or awk as described by @Kusalananda and @SriniV, respectively. One problem is that there will always be a trailing , because of the trailing NUL. This can be rectified by a sed command where we print everything but the last comma, sed 's/\(.*\),/\1/'. Note that the .* will print everything until the last comma. The \1 prints everything in the (backslash escaped) parentheses, i.e. \1 will print whatever was found by the .* part of \(.*\),
让我们用 tr
$ id --zero -Gn $FT_USER | tr '\0' ',' | sed 's/\(.*\),/\1/'
Users,CONSOLE LOGON,Authenticated Users,CurrentSession,LOCAL
现在，我们将使用 awk
$ id --zero -Gn $FT_USER | gawk -F "\0" '{$1=$1}1' OFS="," | sed 's/\(.*\),/\1/'
Users,CONSOLE LOGON,Authenticated Users,CurrentSession,LOCAL
编辑
如果你(和我一样)觉得最后的 sed 有点作弊，这里有一个替代方案:
Edit
If you feel (as I do) that the sed at the end is a bit of a cheat, here's an alternative:
$ id -Gnz $FT_USER | awk 'BEGIN{FS="\0"; OFS=","}{NF--; print}'
Users,CONSOLE LOGON,Authenticated Users,CurrentSession,LOCAL
另请注意，无论您在哪里看到上面的 --zero -Gn，都可以将其替换为 -Gnz.-G 打印出当前用户所属的所有组，-n 打印名称，而不是组 ID (GID)，-z 做我们在上面描述的 --zero 做的事情
Also note that, wherever you see --zero -Gn above, you can replace it with -Gnz. -G prints out all of the groups to which the current user belongs, -n prints the name, rather than the group id (GID), and -z does what we described --zero doing above
这篇关于来自 $FT_USER 的逗号分隔列表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！