How do I prevent echo eating spaces?

[essdeeay]

Hello

I'm writing a script in which I need to insert date/time stamps in exactly the same format as syslog (ie. %b %e %T). However, echo is eating the blank padded space for (%e).

I will insert the date/time stamp like this:

echo $DATE This is a date time stamp

and so have defined the DATE variable like this:

DATE=`date '+%b %e %T'`

This is where echo eats up the blank padded space.

If I were to insert the date/time stamp using "$DATE" instead of just $DATE, echo does not eat the spaces, but I don't want to do that and can't figure out how to include this in the actual DATE variable.

If I use sed (it's getting dirty and cumbersome now), I can do this (using echo -E $DATE):

DATE=`date '+%b %e %T' | sed 's/ /*/g'`
Nov**1*19:21:27

Which works as expected, but when I try and use an escaped space instead of an asterisk in the replacement text, it just won't work. Below, what I think should be being fed back to the echo command is:

Nov\ \ 1\ 19:21:27

but this is what I get instead:

DATE=`date '+%b %e %T' | sed 's/ /\\ /g'`
Nov 1 19:21:27

Any help would be appreciated.

Many thanks,
Steve

[Firewing1]

You have to use quotes:
...
echo "$DATE This is a date time stamp"
Firewing1

[essdeeay]

Thanks for your reply Firewing1

I was hoping to not have to use the " " to preserve the variable, because AFAIK you normally don't have to. However, it seems " " is the answer so that's the way I shall go.

Again, many thanks!
Steve

[Firewing1]

Yeah... BASH I found isn't the best language for programming. Stuff like this makes it annoying. I much rather PHP.
Firewing1