在模式中使用awk命令的参数时出错

Question

I'm running these commands:

ps -ef | awk -v piddd="$child_pid" '$2 ~ /\<piddd\>/ { print $3; }'
ps -ef | awk -v piddd="$child_pid" '$2 ~ /piddd/ { print $3; }'

It doesn't give me any result. When I try with this one, I get what I need, although in some cases I'll get additional pids:

ps -ef | awk -v piddd="$child_pid" '$2 ~ piddd { print $3; }'

What is wrong with first ones?

Answer 1

You cannot use a variable with the /pattern/ syntax.

If you want to add word boundaries (and your version of awk supports the syntax), you can do so by concatenating strings:

ps -ef | awk -v piddd="$child_pid" '$2 ~ "\\<" piddd "\\>" { print $3 }'

Note that the \\ must be escaped in this case.

If you just want the whole field to match the exact variable, I'd suggest using a simple string comparison:

ps -ef | awk -v piddd="$child_pid" '$2 == piddd { print $3 }'

Answer 2

There isn't a need to parse the output of ps utility just to get the parent pid PPID of a child PID . The ps utility already provides this functionality.

ps -o ppid= -p $child_pid

The parameter -o ppid= tells ps to just print the parent pid. Without = the printout will contain a header PPID .

The parameter -p $child_pid tells ps to get the process information from the process id identified by variable $child_pid .

Answer 3

ps -ef | awk '$2 ~ /'`echo "$child_pid"`'/ { print $3; }'

# or

ps -ef | awk '$2 ~ /'$child_pid'/ { print $3; }'

# or create function
function father_pid() { ps -ef | awk '$2 ~ /'$1'/ { print $3; }'; }

# use function
father_pid $child_pid