LinuxQuestions.org - [SOLVED] get n-variable parameters in bash

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get n-variable parameters in bash

Hi,

I've a script that it's invoked with n-variable parameters. Here's an examples:

Code:

./myprogram.sh inputdir FIELD1 FIELD2 ... FIELDN outputfile

In the script I would like to get the FIELD names that were passed.

How can I do that in bash?

You can try the shift statement inside a loop to cycle through all the positional parameters (arguments) independently from their number, e.g.

Code:

until [[ -z $1 ]]

do

  echo $1

  shift

done

the loop is executed until $1 is null. Please, see the Bash Reference Manual for more details.

I've found it.
The command getopt worked for me.

Given a command line:

Code:

prompt$ verb option1 option2 ... optionN

where is token is a group of characters separated by whitespace(blanks, tabs, etc)

You access them within your bash script as:

$0 -- the name of the "verb" (see more below)
$1 $2 ... $9 -- each option in order from left to right
$* -- all of the options as a single value

You may learn more at a terminal with

Code:

prompt$ man bash

and read about "Positional Parameters". NOTE -- 'man' launches the 'less' page reader.
Within less, type "/Positional" to search for that string.

If you have more than nine space-separated options, there is a built-in command, 'shift' to drag more options into place. Again 'man bash' is your friend.

The value of $0 is actually the full file path and name to the running script. So if you are running a script named "framis" that is stored in "/usr/local/bin", the verb will appear as "/usr/local/bin/framis".

Bonne chance,
~~~ 0;-Dan

Quote:

Originally Posted by SaintDanBert

If you have more than nine space-separated options, there is a built-in command, 'shift' to drag more options into place.

Actually this is not required, but of course still works. You can however simply use the following for greater than 9:

Code:

echo ${10}