[Bash] Print out how many number of arguments and arguments

Bash Script :



args=("$@")

Number=("$#")

echo $# arguments passed



for i in $(seq 0 1 $Number)

do

   echo ${args[$i]}

done



Result :


Happy@Happy-laptop:~/Desktop$ bash test123.sh 1 2 3 4 5 6  7  8

8 arguments passed

1

2

3

4

5

6

7

8



Reference :

• In Bash, how do you access command line arguments inside a function?
• Bash For Loop Examples