UNIX for Dummies Questions & Answers

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.

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

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.