CODEWITHSUNDEEP
basharguments
sabri
sabri

Reputation: 2309

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

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

Are there other (better) ways of doing this?

Upvotes: 228

Views: 252259

Answers (4)

zsalzbank
zsalzbank

Reputation: 9857

The number of arguments is $#

Search for it on this page to learn more.

Upvotes: 356

VIPIN KUMAR
VIPIN KUMAR

Reputation: 3137

Below is the easy one -

cat countvariable.sh

echo "$@" | awk '{print NF}'

Output :

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

Upvotes: -3

Michael Brux
Michael Brux

Reputation: 4406

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."

Upvotes: 24

Dennis Williamson
Dennis Williamson

Reputation: 360095

#!/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"

Upvotes: 112

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