使用grep过滤进程(Using grep to filter process)

使用grep过滤进程(Using grep to filter process) root 1626 1 0 04:49 ? 00:00:00 /usr/sbin/sshd USERNAME 5133 5038 0 14:12 pts/0 00:00:00 /bin/bash ./find_proc.sh sshd USERNAME 5137 5133 0 14:12 pts/0 00:00:00 grep sshd

我如何使用grep过滤底部进程？

ps -ef | grep "$1"

是什么让我输出。 我知道你需要使用grep -v来过滤它，但是我不确定它的顺序。它需要一个$ 1的Argument。 所以我在终端的输入是./script_file sshd来获取上面的输出。 有什么建议么？

ps -ef | grep "$1" | grep -v grep

没有输出。

root 1626 1 0 04:49 ? 00:00:00 /usr/sbin/sshd USERNAME 5133 5038 0 14:12 pts/0 00:00:00 /bin/bash ./find_proc.sh sshd USERNAME 5137 5133 0 14:12 pts/0 00:00:00 grep sshd

How can i use grep to filter out the bottom to processes?

ps -ef | grep "$1"

is what gets me that output. I know you need to use grep -v to filter it out, however i am not sure of the order to do it in. It takes an Argument which is $1. So my input in the terminal is ./script_file sshd to get the output above. Any suggestions?

ps -ef | grep "$1" | grep -v grep

gives me no output.

最满意答案

你可以使用pgrep 。 大多数现代Linux和Unix系统都可以使用它（在home brew OSX上可用）：

pgrep -fl "$1"

You can use pgrep. It is available on most of the modern Linux and Unix systems (on OSX available via home brew):

pgrep -fl "$1"