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Unable to store "-e" in variable ??????

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# 1  
Old 08-13-2010
Error Unable to store "-e" in variable ??????
p="-e"
echo $p

It is not returning the value "-e" stored. Instead returns null.

I am wondering how could this happen. Please help me out.I tried all possibilities like p='-e' | p="\-e". Nothing seems to work.

SmilieSmilie
# 2  
Old 08-13-2010
echo assumes the -e is a commandline parameter(echo does indeed take -e on some systems, see man echo) and uses it instead of echoing it. To tell echo that the following string contains no parameters, tell it echo --.

Code:
STR="-e"
echo -- "$STR"

This User Gave Thanks to Corona688 For This Post:
shanneykar
# 3  
Old 08-13-2010
Thanks corona. It works, but still

echo -- $str ==> -- -e

Is it posible to give just teh stored value "-e" or so we need to cut and get the variable.?

---------- Post updated at 04:47 PM ---------- Previous update was at 04:29 PM ----------

I finally settled for :

str="-e"

echo .$str | sed 's/\.//1'

-e

So wherever I want to echo variables (even if it contains -e/-n/-E ) I used the above & it works fine. Thanks corona !! Smilie
# 4  
Old 08-13-2010
Code:
printf %s\\n "$p"

This User Gave Thanks to alister For This Post:
shanneykar
# 5  
Old 08-13-2010
Cool.. Wat %s mean? and will it be issue to use printf in ksh?
# 6  
Old 08-13-2010
"%s" means "substitute a string". See man printf. printf is an external command that should be present for any shell.
# 7  
Old 08-13-2010
Corona,

I need assistance on another issue too.
I want a variable to be left indented with fixed size.

With "typeset -Z100 temp" am able to get fixed size variable, but it is rigt indented.
Any Idea to get left indented?
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