How do I take optional parameters in my shell script.
Command Line argument would be: ./abc.sh parm1 parm2 parm3 parm4
Here parm3 and parm4 are optional depending upon parm1(config)
How to write a shell script which will take mandatorily parm1 and parm2 as argument and treat the other two as optional parameters
Use ${var?} syntax for required parameters:
#!/bin/sh : ${1:?first argument is required} : ${2:?second argument is required} You can check the number of argument with $#, so you can do things like:
#!/bin/sh die() { echo "$@" >&2; exit 1; } case ${1?} in foo) test $# = 4 || die When first argument is foo, you must give 4 args;; bar) test $# = 2 || die When first argument is bar, only give 2 args;; *) ;; esac 
test -n "$3" will pass if the empty string is passed as an argument, so checking $# is the only way to handle optional values that could be empty.
: ${1?first arg is required} to display an error only if the parameter is missing altogether. Use : ${1:?...} if a null string is not permitted as the value of the argument.
test -n "$3" fails when $3 is the empty string, and I'm making the tacit assumption that the empty string is unacceptable throughout.
./abc.sh p1 p2 p3 p4 p5if, as your question indicates, only 2, 3 or 4 arguments are acceptable. You should know about$#, the special parameter (variable) that reports the number 'command line arguments' ('positional parameters' in the formal jargon).