How to replace specific row using sed or awk based with run command

awk -F, -v q=\' '
  function shellquote(s) {
    gsub(q, q "\\" q q, s)
    return q s q
  }
  $4 == "" {
    printf "%s,", $1
    system("cmd " shellquote($1))
    next
  }
  {print}'

The shellquote part is important. Consider the case of a value of $1 like ;reboot for instance.

You could use the following code:

awk -v output1="$(cmd1)" 'BEGIN{FS=OFS=","} NF&&NF<4{print $1,output1}1' file

The option -v set the variable output1 as the result of the shell command cmd1.

The statement BEGIN{FS=OFS=","} is setting the delimiter to the comma.

If the line is not empty and the number of fields is lower than 4, replace the line.

At last, the 1 triggers the default statement in awk: print the whole line.

awk -F, '{ if ( $4 == "" ){printf "%s,",$1 ; system(CMD) ;}else {print $0}}' test.csv

If the fourth filed is empty print the first field with the output of the command, else print the hole line.

But be aware of the output of the command how it will print

may be this post helps you http://stackoverflow.com/questions/20646819/how-can-i-pass-variables-from-awk-to-a-shell-command

$ cat input.txt
a,b,c,d
a,b
a,b,c,d
a,b,c
a,b,c,d


$ awk -F, 'NF<4{cmd="echo output1";while(cmd|getline line){printf("%s,%s\n",$0,line)};close(cmd);next}1' input.txt
a,b,c,d
a,b,output1
a,b,c,d
a,b,c,output1
a,b,c,d