How do I find the number of arguments passed to a Bash script?
The number of arguments is $#
Search for it on this page to learn more: http://tldp.org/LDP/abs/html/internalvariables.html#ARGLIST
#!/bin/bash
echo "The number of arguments is: $#"
a=${@}
echo "The total length of all arguments is: ${#a}: "
count=0
for var in "$@"
do
echo "The length of argument '$var' is: ${#var}"
(( count++ ))
(( accum += ${#var} ))
done
echo "The counted number of arguments is: $count"
echo "The accumulated length of all arguments is: $accum"
to add the original reference:
You can get the number of arguments from the special parameter $#
. Value of 0 means "no arguments". $#
is read-only.
When used in conjunction with shift
for argument processing, the special parameter $#
is decremented each time Bash Builtin shift
is executed.
see Bash Reference Manual in section 3.4.2 Special Parameters:
"The shell treats several parameters specially. These parameters may only be referenced"
and in this section for keyword $# "Expands to the number of positional parameters in decimal."