Problem to get arguments with ${var}

philipina · 07-05-2004, 03:13 AM

Hello,

I would like to use a FOR statement to display all arguments received by my script.

I'm using the following code:
-------------------------------------
#!/bin/bash

declare nArguments=$#

for (( i=1; i<$nArguments; i++ ));
do
echo "var = ${i}"
done
-------------------------------------

echo "var = ${i}" --> this is not working, I also tried echo "var = ${\$i}" but this is the same result.

Thanks in advance for you help.

Alain.

hw-tph · 07-05-2004, 04:42 AM

I am not sure the C/Java-like for loop you use in your example actually works in sh/bash. You could use a for loop without the "in" part (usually you would type "for item in list do...") to iterate through a list of commandline arguments:

Code:

#!/bin/bash
declare nArguments=$#

echo "nArguments=$nArguments"

for a
do
	echo "$a "
done

philipina · 07-05-2004, 04:58 AM

No, this is not working.

Thank.

Alain.

Dark_Helmet · 07-05-2004, 05:36 AM

Code:

#!/bin/bash

declare nArguments=$#

echo "nArguments=${nArguments}"

current_argument=${1}
argument_number=1
while [ ! -z "${current_argument}" ]
do
  echo "Argument number ${argument_number}: ${current_argument}"
  shift
  current_argument=${1}
  ((argument_number=argument_number+1))
done

exit 0

As hw-tph said, I don't believe C-style for loops wok in bash; nor do variable++ type statements. man bash for details.

philipina · 07-05-2004, 06:10 AM

Thank it works.

This "shift" command is very useful for me, I didn't know it before!

Thank again.