在 Shell 脚本中获取进程的 PID

Question

I am writing one shell script and I want to get PID of one process with name as "ABCD". What i did was :

process_id=`/bin/ps -fu $USER|grep "ABCD"|awk '{print $2}'`

This gets PID of two processes ie of process ABCD and the GREP command itself what if I don't want to get PID of GREP executed and I want PID only of ABCD process?

Please suggest.

Answer 1

只需 grep 即可 grep 本身！

process_id=`/bin/ps -fu $USER| grep "ABCD" | grep -v "grep" | awk '{print $2}'`

Answer 2

您是否尝试过使用pidof ABCD ？

Answer 3

It's very straight forward. ABCD should be replaced by your process name.

#!/bin/bash

processId=$(ps -ef | grep 'ABCD' | grep -v 'grep' | awk '{ printf $2 }')
echo $processId

Sometimes you need to replace ABCD by software name. Example - if you run a java program like java -jar TestJar.jar & then you need to replace ABCD by TestJar.jar .

Answer 4

ps 有一个选项：

process_id=`/bin/ps -C ABCD -o pid=`

Answer 5

You can also do away with grep and use only awk .
Use awk 's expression matching to match the process name but not itself.

/bin/ps -fu $USER | awk '/ABCD/ && !/awk/ {print $2}'

Answer 6

You can use this command to grep the pid of a particular process & echo $b to print pid of any running process:

b=`ps -ef | grep [A]BCD | awk '{ printf $2 }'`
echo $b

Answer 7

ps | pgrep ABCD

可以尝试上面的命令返回ABCD进程的进程id。

Answer 8

我找到了一个更好的方法来做到这一点。

top -n 1 | grep "@#" | grep -Eo '^[^ ]+'