variable substitution removing quotes

Question

I seem to have some difficulty getting what I want to work. Basically, I have a series of variables that are assigned strings with some quotes and \ characters. I want to remove the quotes to embed them inside a json doc, since json hates quotes using python dump methods.

I figured it would be easy. Just determine how to remove the characters easy and then write a simple for loop for the variable substitution, well it didn't work that way.

Here is what I want to do.

There is a variable called "MESSAGE23", it contains the following "com.centrify.tokend.cac", I want to strip out the quotes, which to me is easy, a simple echo $opt | sed "s/\"//g". When I do this from the command line:

$> MESSAGE23="com."apple".cacng.tokend is present"
$> MESSAGE23=`echo $MESSAGE23 | sed "s/\"//g"`
$> com.apple.cacng.tokend is present

This works. I get the properly formatted string.

When I then try to throw this into a loop, all hell breaks loose.

for i to {1..25}; do
   MESSAGE$i=`echo $MESSAGE$i | sed "s/\"//g"`
done

This doesn't work (either it throws a bunch of indexes out or nothing), and I'm pretty sure I just don't know enough about arg or eval or other bash substitution variables.

But basically I want to do this for another set of variables with the same problems, where I strip out the quotes and incidentally the "\" too.

Any help would be greatly appreciated.

score 3 · Answer 1 · answered Mar 28 '13 at 00:19

3

You can't do that. You could make it work using eval, but that introduces another level of quoting you have to worry about. Is there some reason you can't use an array?

MESSAGE=("this is MESSAGE[0]" "this is MESSAGE[1]")
MESSAGE[2]="I can add more, too!"
for (( i=0; i<${#MESSAGE[@]}; ++i )); do
   echo "${MESSAGE[i]}"
done

Otherwise you need something like this:

 eval 'echo "$MESSAGE'"$i"'"'

and it just gets worse from there.

answered Mar 28 '13 at 00:19

Mark Reed

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yuck.. no, not opposed to using an array? I have variables MESSAGE1-25, Should I push each of those onto an array and then pop each off and substitute it? I guess I'm not sure where I should put this array? – HM Stanley Mar 28 '13 at 00:37
oh.. I get it. MESSAGE is the array and pop values onto that array – HM Stanley Mar 28 '13 at 00:37
Right. Instead of `MESSAGE1=blah1`, `MESSAGE2=blah2`, etc, you can do `MESSAGE[0]=blah1`, `MESSAGE[1]=blah2`, etc. Or initialize the whole thing with parentheses as I did, or read it in from a file using `read -a` - it just depends on how those variables are getting set in your application. – Mark Reed Mar 28 '13 at 14:13

score 1 · Answer 2 · answered Mar 28 '13 at 00:48

First, a couple of preliminary problems: MESSAGE23="com."apple".cacng.tokend is present" will not embed double-quotes in the variable value, use MESSAGE23="com.\"apple\".cacng.tokend is present" or MESSAGE23='com."apple".cacng.tokend is present' instead. Second, you should almost always put double-quotes around variable expansions (e.g. echo "$MESSAGE23") to prevent parsing oddities.

Now, the real problems: the shell doesn't allow variable substitution on the left side of an assignment (i.e. MESSAGE$i=something won't work). Fortunately, it does allow this in a declare statement, so you can use that instead. Also, when the sees $MESSAGE$i it replaces it will the value of $MESSAGE followed by the value of $i; for this you need to use indirect expansion (`${!metavariable}').

for i in {1..25}; do
    varname="MESSAGE$i"
    declare $varname="$(echo "${!varname}" | tr -d '"')"
done

(Note that I also used tr instead of sed, but that's just my personal preference.)

(Also, note that @Mark Reed's suggestion of an array is really the better way to do this sort of thing.)

Thanks guys.. Learned a ton here from both of you, will experiment with both the array option and using declare. Thanks — HM Stanley, Mar 29 '13 at 01:31

variable substitution removing quotes

2 Answers2