Shell – Using variable with awk -v in a shell script

awkshell-script

I am modifying/re-writing some of the scripts written by former employees of my company and in one of the shell scripts I found the following line.

b=`benchmark=30;grep "Waiting for timer" wk.txt | awk -vbenchmark=$benchmark '$6 > benchmark' | wc -l`

But my bad, I couldn't figure out what the above line is trying to do. I am confused about the usage of 'benchmark' variable here. I created a dummy 'wk.txt' file with the following values and the when echoed, variable $b got the value 1 (which is just a line count of the output of the grep+awk command.

[sreeraj@server ~]$ cat wk.txt
24  here  above the Waiting for timer 37 make sure

Could someone explain what the script author is trying to do with the $benchmark?

awk man page says the below for -v, but I am not I understood what it does.

-v var=val
--assign var=val
          Assign the value val to the variable var, before execution of the program begins.   Such  variable  values  are
          available to the BEGIN block of an AWK program.

Best Answer

If you do:

awk -v benchmark=30 '...'

That is the same as:

awk 'BEGIN{ benchmark = 30 } ... '

This is used to set an initial value for that variable.

Though I don't see why the author does:

benchmark=30; ... | awk -v benchmark=$benchmark ..

They might as well do:

... | awk -v benchmark=30 ..

Related Solutions

Shell Script – Assign Every Word from a Line to a Variable

You can do this with a while read loop like so:

while read -r col1 col2 col3 col4 trash; do
    something with "$col1"
    something with "$col2"
    something with "$col3"
    something with "$col4"
done < /path/to/quaternary_splitted.csv

This will read through each line of quaternary_splitted.csv and set the first column to col1, second column to col2, etc.

The trash parameter is used to catch anything that may be in your file and unwanted. Say you had a line: Nb Ta Cr W what is this doing here?. Without trash you would get:

col1=Nb
col2=Ta
col3=Cr
col4='W what is this doing here?`

linux – Why Does Variable Assigned ZERO Have Value TWO?

It is clear from you edit that the second script is called enigma2.sh, and that you call it by writing enigma2.sh on the command line. The script is somewhere in your path (you declare that it is at /usr/bin/enigma2.sh, and I shall assume that /usr/bin is in your PATH variable). Combining that the script could be found (in the PATH) by the shell and that the first line of the script is #!/bin/sh, the kernel will start a new process that has a command line of:

/bin/sh enigma2.sh

You can confirm this by simply making the first lines of that script as this:

#!/bin/sh 
ps ax|grep enigma2
exit 0
...

You don't need to erase the script, just place the lines shown above at the start of the script and leave all the rest below.

Run it to get:

$ enigma2.sh
 3109 pts/6    S+     0:00 /bin/sh enigma2.sh
 3111 pts/6    S+     0:00 grep enigma2

There are two processes that include the word enigma2.

That explains part of the problem. Now, if you change the script to:

#!/bin/sh 
enigma_pid=$(ps ax|grep enigma2)
echo "$enigma_pid"
exit 0
...                     # leave the old script below (no need to erase it).

And run it:

$ enigma2.sh
 3509 pts/6    S+     0:00 /bin/sh ./enigma2.sh
 3510 pts/6    S+     0:00 /bin/sh ./enigma2.sh
 3512 pts/6    S+     0:00 grep enigma2

When you use a $(…) one additional process appear.

Removing the grep process with grep -v grep leaves 2 processes.

That is the number you get 2.

Note: Please, don't use uppercase variable names, those are reserved for environment variables.
Don't use ~~ENIGMA_PID~~

Solutions

First: If you are going to identify a process by its name, don't use the exact same name in other programs or scripts. You may call the script enigma33.sh.
Second: In this case, you can restrict grep to match a name that ends at 2. The script ends in .sh and won't be counted. Use
```
 grep '[e]nigma2$'
```

And, since all you need is the count of matching lines use (-c):

enigma_pid=$(ps ax | grep -c '[e]nigma2$')

Yes, no need for grep -v grep since the regex [e]nigma2$ can not be equal (as an string) to enigma2.

Edited script:

#!/bin/sh


# Detect If enigma2 is running.
enigma_pid=$(ps ax|grep -c '[e]nigma2$')
echo "$enigma_pid"    

# Deetct if minisatip is running.
mini_pid=$(ps ax|grep -c '[m]inisatip$' )


# Check if the no_enigma file exists
if [ -f /home/root/no_enigma ]; then
    if [ -z "${enigma_pid##*[!0-9]*}" ] ; then
       echo "error: Not a number" >&2; exit 1
    else
        echo "It IS a number"
    fi
    
    if [ "$enigma_pid" -gt 0 ]; then
        echo "$enigma_pid"
        echo "enigma2 is running"
        killall -r '[e]nigma2$'           # make killing more restrictive
    else
        echo "enigma2 is not running"
    fi
    # Check if minisatip is running already, if not then start it
    if [ "$mini_pid" -gt 0 ]; then
        echo "minisatip is already running"
    else
        echo "minisatip is not running"
        echo "starting minisatip"
        /usr/bin/minisatip --satip-xml http://127.0.0.1:8554 -R /usr/share/minisatip/html       
    fi
else    
    # The no_enigma file does not exist
    # Check if minisatip is running, if yes then kill it
    if [ "$mini_pid" -gt 0 ]; then
        echo "minisatip is running, killing it now"
        killall minisatip
    else
        echo "minisatip is not running"
    fi  
    # Check if enigma2 is running already, if not then start it
    if [ "$enigma_pid" -gt 0 ]; then
        echo "enigma2 is already running"
    else
        echo "enigma2 is not running"
        echo "starting enigma2"
        /home/root/run_enigma.sh            
    fi
fi