Linux 管道输出切割

Piping output to cut

提问by ZPS

I am trying to get the name of the shell executing a script.

我正在尝试获取执行脚本的 shell 的名称。

Why does

为什么

echo $(ps | grep $PPID) | cut -d" " -f4

work while

工作的同时

echo ps | grep $PPID | cut -d" " -f4

does not?

才不是？

采纳答案by ruakh

The reason is that

原因是

echo ps

just prints out the string ps; it doesn't run the program ps. The corrected version of your command would be:

只是打印出字符串ps；它不运行程序ps。您的命令的更正版本将是：

ps | grep $PPID | cut -d" " -f4

Edited to add:paxdiablo points out that ps | grep $PPIDincludes a lot of whitespace that will get collapsed by echo $(ps | grep $PPID)(since the result of $(...), when it's not in double-quotes, is split by whitespace into separate arguments, and then echooutputs all of its arguments separated by spaces). To address this, you can use trto "squeeze" repeated spaces:

编辑添加：paxdiablo 指出ps | grep $PPID包含大量空格，这些空格将被折叠echo $(ps | grep $PPID)(因为 的结果$(...)，当它不在双引号中时，被空格分割成单独的参数，然后echo输出所有由空格分隔的参数）。为了解决这个问题，您可以使用tr“挤压”重复空格：

ps | grep $PPID | tr -s ' ' | cut -d' ' -f5

or you can just stick with what you had to begin with. :-)

或者你可以坚持你必须开始的东西。:-)

回答by sarnold

I think if you run just echo pson its own, you'll see why your pipe doesn't work:

我想如果你echo ps自己运行，你会明白为什么你的管道不起作用：

$ echo ps
ps

Instead, check $0. Note that it might be -bashor bash, depending if it is a login shell. (Well, any shell -- not just bash-- but the hyphenis added if the shell is a login shell.)

相反，检查$0. 请注意，它可能是-bash或bash，具体取决于它是否是登录 shell。（好吧，任何外壳——不仅仅是bash——但如果外壳是登录外壳，则会添加连字符。）

回答by icyrock.com

The first line:

第一行：

echo $(ps | grep $PPID) | cut -d" " -f4

says:

说：

• Execute ps | grep $PPIDin a sub-shell

• Return the output - which will be something like this:

  3559 pts/1 00:00:00 bash
  

  and then use that output as the first parameter of echo - which, in practice, means just echo the output

• Then run cut -d" " -f4on that - which gives you the command name in this case

• ps | grep $PPID在子shell中执行

• 返回输出 - 这将是这样的：

  3559 pts/1 00:00:00 bash
  

  然后将该输出用作 echo 的第一个参数 - 实际上，这意味着只回显输出

• 然后运行cut -d" " -f4它 - 在这种情况下它会为您提供命令名称

The second command:

第二条命令：

echo ps | grep $PPID | cut -d" " -f4

says:

说：

• Echo string ps
• Grep that string for $PPID- this will never return anything, as $PPIDcontains a number, so it will never be ps. Thus, grep returns nothing
• Execute cut -d" " -f4with input of the previous command - which is nothing, so you get nothing

• 回声串 ps
• Grep 该字符串 for $PPID- 这将永远不会返回任何内容，因为$PPID包含一个数字，所以它永远不会是ps. 因此，grep 不返回任何内容
• 执行cut -d" " -f4与前一个命令的输入-这是什么，所以你什么也得不到

回答by CognizantApe

For many uses it is better to just us tr. For example:

对于许多用途，最好只使用我们tr。例如：

Input: echo "field1,field2,field3" | tr , \\n Output: field1 field2 field3

输入： echo "field1,field2,field3" | tr , \\n 输出： field1 field2 field3