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linuxbashawksedps

Using awk to get a specific string in line

I have a java process containing multiple strings in the ps output. I just want a particular string out of it.

For example, I have 

root 5565 7687  0 Nov20 ?  00:00:54 /bin/java -Xms256m -Xmx512m -XX:MaxPermSize=128m  -XX:PermSize=256m -XX:MaxPermSize=256m -Dstdout.file=/tmp/std.out -da -Dext.dir.class=profiles/ -Dprocess.name=java1 -Djava.security.policy=security.policy

I want just want process.name=java1 and require nothing else from the ps output. I was unable to find a tangible way to do so using awk or sed.

I am tried to use:

ps -ef | grep java1 | grep -v grep | awk '/process.name/ {print $0}'

The output I get is the ps -ef out.

Is there a simpler way?

Solution

  • Here is one way:

    ps -ef | awk -F"process.name" '{split($2,a," ");print FS a[1]}'
process.name=java1