Very simple argument in scripting question - UNIX for Dummies Questions & Answers

eeto · #1 03-26-2007

I tried to do a search, but it couldnt pinpoint what my answer since using limited but broad keywords. Sorry in advance ; ;

Im limited to using Bourne shell scripting only, atm I have the following code (just the heading part of it)

Code:

...
...
# VARIABLE DECLARATION
# ====================

lineCount=1
eofFlag=0
matched=0

# Ask for player name

echo -n "Player name? "       # < problems with this
# and store it to the variable
read srcStr
...
...

read -n1 keyPress                # < problems with this

...
...

first my echo is showing the -n, and second my read is giving me "is not an identifier" error msg. I tried both echo -n and read -n1 in plain shell environment (non-scripting) and it works fine. But why it wont work when i put them in scripting file. And what are my alternatives?

Thanks for any help.... Couldnt find any answer to my question in google, and I'm stuck and desperate.

Raghuram.P · #2 03-26-2007

Hi,
If possible can you post the full source....atleast from first line of ur script till the line where you get the first error..
I can take a look into it.

Thanks
Raghuram

Dhruva · #3 03-26-2007

yes as raghram said pasting full source will help.
The echo is printing -n on screen means you dont have -n option available so you can use \c(escape seq) to solve the same prob.

Code:

echo "this is test\c"
read x

will do same as

Code:

echo -n "this is test"
read x

pls note read -n option is available with ksh shell and echo -n is with csh.

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