parsing standard input with bash

arosales · 07-15-2005, 11:10 AM

Hi! I'm trying to set a variable with the standard input but i have a problem. For example:

user@work005:~> input some command done by me with ten parameters or more

In mi bash script I have
....
field1=$1
field2=$2 ...
field10=$10
--------------------

fields 1 to 9 i don't have troubles, but field 10 sets "$field1"0, it means

....
field1=input
field2=some...
field10=input0
----------------------

and I need filed10=or

What should I do? Thanks in advance...

arosales · 07-15-2005, 11:15 AM

I just have to enclosed it in {brackets}

field10=${10}

field10=or

(I have to read more before post)

macemoneta · 07-15-2005, 11:26 AM

If the number of positional parameters ($#) is greater than 9, then you can use the 'shift' bash command to get the remaining parameters. See 'man bash' for details. For example:

field1=$1
field2=$2...
field9=$9
shift 9
field10=$1
field11=$2...

eddiebaby1023 · 07-16-2005, 10:42 AM

Quote:

Originally posted by arosales
Hi! I'm trying to set a variable with the standard input but i have a problem.

To be pedantic, you're not setting your variables from standard input, but from command line parameters.