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Escaping embedded variables

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# 1  
Old 01-12-2009
Question Escaping embedded variables
I'm running into a problem with a differential backup script written in GNU Bash 3.0 - the following stripped down code demonstrates the problem quite nicely.

Code:
$ DATE="last tuesday"
$ date --date="$DATE"
Tue Jan  6 00:00:00 PST 2009

So far so good.

Code:
$ CMD="date --date=\"$DATE\""
$ echo $CMD
date --date="last tuesday"
$ $CMD
date: the argument `tuesday"' lacks a leading `+';
When using an option to specify date(s), any non-option
argument must be a format string beginning with `+'.
Try `date --help' for more information.

How can I wrap these variables up correctly? It seems like it's not parsing the quotes correctly, but from the contents of the CMD variable (see above), it looks fine. What gives?
# 2  
Old 01-12-2009
Code:
DATE="last tuesday"
date --date="+$DATE"

# 3  
Old 01-12-2009
Quote:
Originally Posted by Ikon
Code:
DATE="last tuesday"
date --date="+$DATE"

Code:
$ DATE="last tuesday"
$ CMD="date --date=\"+$DATE\""
$ $CMD
date: the argument `tuesday"' lacks a leading `+';
When using an option to specify date(s), any non-option
argument must be a format string beginning with `+'.
Try `date --help' for more information.

No dice.
# 4  
Old 01-12-2009
Code:
# DATE="last tuesday"
# CMD=`date --date="+$DATE"`
# echo $CMD
Tue Jan 6 00:00:00 MST 2009

# 5  
Old 01-12-2009
That just runs "date" immediately and sticks the output into the variable CMD. I want to be able to craft a longer command out of several variables, and then run it multiple times.
# 6  
Old 01-12-2009
Code:
# DATE="last tuesday"
# CMD="date --date=\"+$DATE\""
# eval $CMD

# 7  
Old 01-12-2009
Hah! That did it - thanks Ikon!!
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