How do I find the number of arguments passed to a Bash script?

Question

How do I find the number of arguments passed to a Bash script?

This is what I have currently:

#!/bin/bash
i=0
for var in \"$@\"
do
  i=i+1
done


        

Answer 1

Below is the easy one -

cat countvariable.sh

echo "$@" |awk '{for(i=0;i<=NF;i++); print i-1 }'

Output :

#./countvariable.sh 1 2 3 4 5 6
6
#./countvariable.sh 1 2 3 4 5 6 apple orange
8

Answer 2

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."

Answer 3

that value is contained in the variable $#

Answer 4

The number of arguments is $#

Search for it on this page to learn more: http://tldp.org/LDP/abs/html/internalvariables.html#ARGLIST

Answer 5

#!/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"