Is there any way to specify a field delimiter for more spaces with the cut command? (like " "+) ?
For example: In the following string, I like to reach value '3744', what field delimiter I should say?

有没有办法用CUT命令为更多的空格指定一个字段分隔符？(如“”+)？例如：在下面的字符串中，我希望达到值‘3744’，我应该说什么字段分隔符？

$ps axu | grep jboss

jboss     2574  0.0  0.0   3744  1092 ?        S    Aug17   0:00 /bin/sh /usr/java/jboss/bin/run.sh -c example.com -b 0.0.0.0

cut -d' ' is not what I want, for it's only for one single space.
awk is not what I am looking for either, but how to do with 'cut'?

Cut-d‘’不是我想要的，因为它只有一个空间。AWK也不是我想要的，但是怎么处理‘Cut’呢？

thanks.

谢谢。

优秀答案推荐

Actually awk is exactly the tool you should be looking into:

实际上，awk正是您应该研究的工具：

ps axu | grep '[j]boss' | awk '{print $5}'

or you can ditch the grep altogether since awk knows about regular expressions:

或者，您可以完全放弃grep，因为awk了解正则表达式：

ps axu | awk '/[j]boss/ {print $5}'

But if, for some bizarre reason, you really can't use awk, there are other simpler things you can do, like collapse all whitespace to a single space first:

但是，如果出于某种奇怪的原因，您真的不能使用awk，您还可以做其他更简单的事情，比如首先将所有空格折叠到一个空格中：

ps axu | grep '[j]boss' | sed 's/\s\s*/ /g' | cut -d' ' -f5

---

That grep trick, by the way, is a neat way to only get the jboss processes and not the grep jboss one (ditto for the awk variant as well).

顺便说一句，grep技巧是一种巧妙的方法，可以只获取Joss进程，而不获取grep Joss进程(awk变体也是如此)。

The grep process will have a literal grep [j]boss in its process command so will not be caught by the grep itself, which is looking for the character class [j] followed by boss.

grep进程将在其process命令中有一个字面意义的grep [j]boss，因此不会被grep本身捕获，后者正在寻找boss后面的字符类[j]。

This is a nifty way to avoid the | grep xyz | grep -v grep paradigm that some people use.

这是避免某些人使用的|grep xyz|grep-v grep范例的好方法。

awk version is probably the best way to go, but you can also use cut if you firstly squeeze the repeats with tr:

AWK版本可能是最好的方法，但如果你首先用tr挤压重复，你也可以使用Cut：

ps axu | grep jbos[s] | tr -s ' ' | cut -d' ' -f5
#        ^^^^^^^^^^^^   ^^^^^^^^^   ^^^^^^^^^^^^^
#              |            |             |
#              |            |       get 5th field
#              |            |
#              |        squeeze spaces
#              |
#        avoid grep itself to appear in the list

I like to use the tr -s command for this

为此，我喜欢使用tr-S命令

 ps aux | tr -s [:blank:] | cut -d' ' -f3

This squeezes all white spaces down to 1 space. This way telling cut to use a space as a delimiter is honored as expected.

这会将所有空格压缩到1个空格。这种方式告诉cut使用空格作为空格是符合预期的。

I am going to nominate tr -s [:blank:] as the best answer.

我将提名tr-S[：BLACK：]作为最佳答案。

Why do we want to use cut? It has the magic command that says "we want the third field and every field after it, omitting the first two fields"

我们为什么要使用CUT？它有一个神奇的命令：“我们需要第三个字段和其后的每个字段，省略前两个字段。”

cat log | tr -s [:blank:] |cut -d' ' -f 3- 

I do not believe there is an equivalent command for awk or perl split where we do not know how many fields there will be, ie out put the 3rd field through field X.

我不相信Awk或Perl Split有类似的命令，我们不知道会有多少个域，即输出第三个域到域X。

Shorter/simpler solution: use cuts (cut on steroids I wrote)

ps axu | grep '[j]boss' | cuts 4

Note that cuts field indexes are zero-based so 5th field is specified as 4

请注意，Cuts字段索引是从零开始的，因此第5个字段被指定为4

http://arielf.github.io/cuts/

Http://arielf.github.io/cuts/

And even shorter (not using cut at all) is:

更短的(根本不使用Cut)是：

pgrep jboss

One way around this is to go:

绕过这个问题的一种方法是：

$ps axu | grep jboss | sed 's/\s\+/ /g' | cut -d' ' -f3

to replace multiple consecutive spaces with a single one.

用一个空格替换多个连续的空格。

Personally, I tend to use awk for jobs like this. For example:

就我个人而言，我倾向于使用awk来做这样的工作。例如：

ps axu| grep jboss | grep -v grep | awk '{print $5}'

As an alternative, there is always perl:

作为替代方案，总是有Perl：

ps aux | perl -lane 'print $F[3]'

Or, if you want to get all fields starting at field #3 (as stated in one of the answers above):

或者，如果您想要获取从字段#3开始的所有字段(如上面的答案之一所述)：

ps aux | perl -lane 'print @F[3 .. scalar @F]'

If you want to pick columns from a ps output, any reason to not use -o?

如果您想从ps输出中选择列，那么有什么理由不使用-o呢？

e.g.

例如

ps ax -o pid,vsz
ps ax -o pid,cmd

Minimum column width allocated, no padding, only single space field separator.

分配的最小列宽，没有填充，只有一个空格字段分隔符。

ps ax --no-headers -o pid:1,vsz:1,cmd

3443 24600 -bash
8419 0 [xfsalloc]
8420 0 [xfs_mru_cache]
8602 489316 /usr/sbin/apache2 -k start
12821 497240 /usr/sbin/apache2 -k start
12824 497132 /usr/sbin/apache2 -k start

Pid and vsz given 10 char width, 1 space field separator.

PID和VSZ提供10个字符宽度，1个空域分隔符。

ps ax --no-headers -o pid:10,vsz:10,cmd

  3443      24600 -bash
  8419          0 [xfsalloc]
  8420          0 [xfs_mru_cache]
  8602     489316 /usr/sbin/apache2 -k start
 12821     497240 /usr/sbin/apache2 -k start
 12824     497132 /usr/sbin/apache2 -k start

Used in a script:-

在脚本中使用：-

oldpid=12824
echo "PID: ${oldpid}"
echo "Command: $(ps -ho cmd ${oldpid})"

I've implemented a patch that adds new -m command-line option to cut(1), which works in the field mode and treats multiple consecutive delimiters as a single delimiter. This basically solves the OP's question in a rather efficient way, by treating several spaces as one delimiter right within cut(1).

我已经实现了一个补丁，它向Cut(1)添加了新的-m命令行选项，它在字段模式下工作，并将多个连续的分隔符视为单个分隔符。这基本上以相当有效的方式解决了OP的问题，将几个空格视为Cut(1)中的一个分隔符。

In particular, with my patch applied, the following command will perform the desired operation. It's as simple as that, just add -m into the invocation of cut(1) and simply use -d ' ' -f 5 to extract the PID values from the process list produced by ps(1):

特别是，在应用我的补丁后，以下命令将执行所需的操作。就这么简单，只需将-m添加到Cut(1)的调用中，然后使用-d‘’-f5从ps(1)生成的进程列表中提取PID值：

ps axu | grep jboss | cut -d ' ' -m -f 5

I also submitted this patch upstream, and let's hope that it will eventually be accepted and merged into the coreutils project.

我也向上游提交了这个补丁，让我们希望它最终会被接受并合并到coreutils项目中。

There are some further thoughts about adding even more whitespace-related features to cut(1), and having some feedback on all that from different people would be great, preferably on the coreutils mailing list. I'm willing to implement more patches for cut(1) and submit them upstream, which would make this utility more versatile and more usable in various real-world scenarios.

还有一些关于在Cut(1)中添加更多与空格相关的特性的进一步想法，从不同的人那里得到一些反馈将是很好的，最好是在coreutils邮件列表上。我愿意为Cut(1)实现更多的补丁，并向上游提交它们，这将使这个实用程序在各种现实场景中更加通用和更有用。

Another way if you must use cut command

另一种方法，如果您必须使用CUT命令

ps axu | grep [j]boss |awk '$1=$1'|cut -d' ' -f5

In Solaris, replace awk with nawk or /usr/xpg4/bin/awk

在Solaris中，将awk替换为nawk或/usr/xpg4/bin/awk

I still like the way Perl handles fields with white space.

First field is $F[0].

我仍然喜欢Perl处理带有空格的字段的方式。第一个字段是$F[0]。

$ ps axu | grep dbus | perl -lane 'print $F[4]'

My approach is to store the PID to a file in /tmp, and to find the right process using the -S option for ssh. That might be a misuse but works for me.

我的方法是将ID存储到/tmp中的一个文件中，并使用ssh的-S选项查找正确的进程。这可能是一种滥用，但对我来说很管用。

#!/bin/bash

TARGET_REDIS=${1:-redis.someserver.com}
PROXY="proxy.somewhere.com"

LOCAL_PORT=${2:-6379}

if [ "$1" == "stop" ] ; then
    kill `cat /tmp/sshTunel${LOCAL_PORT}-pid`
    exit
fi

set -x

ssh -f -i ~/.ssh/aws.pem centos@$PROXY -L $LOCAL_PORT:$TARGET_REDIS:6379 -N -S /tmp/sshTunel$LOCAL_PORT  ## AWS DocService dev, DNS alias
# SSH_PID=$! ## Only works with &
SSH_PID=`ps aux | grep sshTunel${LOCAL_PORT} | grep -v grep | awk '{print $2}'`
echo $SSH_PID > /tmp/sshTunel${LOCAL_PORT}-pid

Better approach might be to query for the SSH_PID right before killing it, since the file might be stale and it would kill a wrong process.

更好的方法可能是在终止SSH_PID之前查询它，因为该文件可能已经过时，并且它会终止错误的进程。