How to write a bash script to label each argument like this:

$ bash argcnt.sh this is a "real live" test

  there are 5 arguments
  arg1: this
  arg2: is
  arg3: a
  arg4: real live
  arg5: test

Cause whatever I do it's only counting as 6 arguments

CodePudding user response：

For one, you are trying to do different formats, resulting in it being an incorrect size. this is a "real live" test

This won't work because you used quotes on a few words, counting that argument as just one argument.

this is a real live test Would work, counting each word as a different argument.

For 2, you are making the code more complicated than it has to be. You are echoing each command individually, instead of making a for loop that does the heavy lifting.

echo "There are $# arguments"

$# tells you the number of arguments there are.

i=1
for arg in $@;
do
    echo -e "arg$i: $arg\n"
    i =1
done

This creates a loop that prints out every argument given and how many arguments there are

CodePudding user response：

You're on the right track. In your example, "real live" is treated as one input. It's in variable $4.

#!/bin/bash  
echo "there are $# arguments"  
echo "arg1: $1" 
echo "arg2: $2" 
echo "arg3: $3" 
echo "arg4: $4" 
echo "arg5: $5" 

When you run this with your input, it should give you the output that you stated.