Unseting a list of dynamic variables in Bash

Question

How can I achieve the following with Bash:

x="search='something something' replace='something' subject=/path/path"

eval "$x"

echo $search
# prints something
# etc.

# (I am trying to do something here)
# So the following wouldn't print **something** anymore

echo $search

I basically want to unset the variables that were set using eval initially.

Fork a subshell, do `eval` in the subshell and once you're done, exit the subshell — anubhava, Dec 04 '17 at 20:06
After `eval`, there's nothing special about the variables. Just use `unset search replace subject`. — chepner, Dec 04 '17 at 20:40
@chepner There are hundreds of them (not sure which will be available) — Adam, Dec 04 '17 at 20:46
That's one reason using `eval` like this is a bad idea. A script shouldn't need to know the *value* of its variables, but it should know which ones are defined. Where is `x` coming from? — chepner, Dec 04 '17 at 20:48

score 3 · Answer 1 · answered Dec 04 '17 at 20:11

3

You can fork a subshell, do eval in the subshell and once you're done, exit the subshell:

# do this in a subshell
(
x="search='something' replace='something' subject=/path"
eval "$x"
echo "search=$search"
)

# now out of subshell
echo "search=$search"

answered Dec 04 '17 at 20:11

anubhava

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Very useful but the script isn't structured in a way to accept this solution. – Adam Dec 04 '17 at 20:33
Can you clarify how do you want it to be structured? – anubhava Dec 04 '17 at 21:07
1

@AdamSilver The script you posted in your question is. Can you update it? – that other guy Dec 04 '17 at 21:07

F. Hauri - Give Up GitHub · Accepted Answer · 2017-12-05T08:55:21.917

3

Preamble: Care using eval as eval is evil...
Whatever...

bash could do this very quickly:

unset ${x//=*([^ ])}

As pointed out by glenn jackman, this work only if there are no spaces in content of variables.

In this case, we could (very quickly too) prepare some intermediary variable (x without spaces):

xwos=${x//\"*([^\"])\"}        # This will drop `search="don't know"`
xwos=${xwos//\'*([^\'])\'}     # This will drop `search='blah blah'`
unset ${xwos//=*([^ ])}

In fine, adding this to your script will do the job:

xwos=${x//\"*([^\"])\"}; xwos=${xwos//\'*([^\'])\'}; unset ${xwos//=*([^ ])}

Explanation / showcase

Of course, you have to use extglob bash's option, if not already set:

shopt -s extglob

Let's try:

x="search='something something' replace='something' subject=/path/path"

eval "$x"

echo $search 
something something

declare -p search replace subject
declare -- search="something something"
declare -- replace="something"
declare -- subject="/path/path"

shopt -s extglob
xwos=${x//\"*([^\"])\"}
printf 'X w/o double quoted spaces 1st:\n    "%s"\n' "$xwos"
X w/o double quoted spaces 1st:
    "search='something something' replace='something' subject=/path/path"

xwos=${xwos//\'*([^\'])\'}
printf 'X w/o single quoted spaces 2nd:\n    "%s"\n' "$xwos"
X w/o single quoted spaces 2nd:
    "search= replace= subject=/path/path"

printf 'String containing variables w/o values:\n    "%s"\n' "${xwos//=*([^ ])}"
String containing variables w/o values:
    "search replace subject"

unset ${xwos//=*([^ ])}

declare -p search replace subject
bash: declare: search: not found
bash: declare: replace: not found
bash: declare: subject: not found

Nota: If you don't need x anymore, you could redefine x instead of populating a new variable:

shopt -s extglob;x=${x//\"*([^\"])\"};x=${x//\'*([^\'])\'};unset ${x//=*([^ ])}

Demo

x="search='something something' replace='something' subject=/path/path"

eval "$x"

echo $search 
something something

shopt -s extglob;x=${x//\"*([^\"])\"};x=${x//\'*([^\'])\'};unset ${x//=*([^ ])}
echo $search

edited Dec 05 '17 at 08:55

answered Dec 04 '17 at 20:55

F. Hauri - Give Up GitHub

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The variable names MUST be extracted from x. I don't know them before hand. – Adam Dec 04 '17 at 21:09
@AdamSilver That's what `${x//=*([^ ])}` does. – PesaThe Dec 04 '17 at 21:10
1

Nice. The only danger is if one of the variable values contains a space. I came up with `grep -oP '\w+(?==)' <<<"$x"` to extract the varnames. Doing it in the shell is much better. – glenn jackman Dec 04 '17 at 21:11
What if search value was 'something something' (with spaces) – Adam Dec 04 '17 at 21:12
@AdamSilver, try this: `vars=$(grep -oP '\w+(?==)' <<<"$x"); echo unset $vars` -- and remove the `echo` if the right varnames were found – glenn jackman Dec 04 '17 at 21:15
@glennjackman I don't get the whole declare parts. Should I have it too? – Adam Dec 04 '17 at 21:20
This approach is good as long as there is a single string being used in `eval`. However if there are multiple then env variable being created from many strings then unsetting like this won't be simple. – anubhava Dec 04 '17 at 21:44
1

@AdamSilver No, the `declare` parts are there just to show it functions correctly. – PesaThe Dec 04 '17 at 21:56
1

@anubhava the entire architecture of the question is flawed. We're all trying to just mitigate the damage – glenn jackman Dec 04 '17 at 22:19
It's not working for me. Here's a full example: https://bpaste.net/show/d98d02b617df – Adam Dec 04 '17 at 22:19
@glennjackman You're right! I've edited my answer, without any fork to `grep`, this is very quick! – F. Hauri - Give Up GitHub Dec 05 '17 at 06:45
@AdamSilver This is tested under many [tag:bash]! Which version are you using (under which OS, distrib)? – F. Hauri - Give Up GitHub Dec 05 '17 at 09:03

Unseting a list of dynamic variables in Bash

2 Answers2

bash could do this very quickly:

Explanation / showcase

Demo