bash scripting - variable substitution and double quoting within a function

Question

So, I'm working on a couple of functions in my bash script. I have one that works, but my 2nd one includes a number of values enclosed in double quotes "", thus breaking my variable substitutions. I've tried surrounding those values with a single tick, but that doesn't work in some cases.

For example:

# Functions
unset_versions ()
{
   host='127.0.0.1'
   db='mydev'
   _account='"foo"'
   _profile='"bar"'
   _mongo=$(which mongo);
   exp="db.profile_versions_20170420.update({account:${_account}, profile:${_profile}, version:${_version}},{ \$unset : { labels : true }});";
   ${_mongo} ${host}/${db} --eval "$exp"
}

However this won't work:

update_versions ()
{
   host='127.0.0.1'
   db='mydev'
   _account='"foo"'
   _profile='"bar"'
   _mongo=$(which mongo);
   exp="db.profile_versions_20170420.update({account:${_account}, profile:${_profile}, version:${_version}},{ \$set : { labels : { "1" : { "name" : "data layer patch", "color" : "regal-rose" }, "2" : { "name" : "Global Functions", "color" : "melon-mambo" }}}});
   ${_mongo} ${host}/${db} --eval "$exp"
}

I know this looks ugly. If there's a better way please provide some suggestions. I also tried doing exp=$(....), but that doesn't seem to work either.

You may want to take a look at this post: http://stackoverflow.com/questions/6697753/difference-between-single-and-double-quotes-in-bash — codeforester, Apr 24 '17 at 04:54

score 1 · Accepted Answer · answered Apr 24 '17 at 03:55

You should escape '"' like this:

exp="db.profile_versions_20170420.update({account:${_account},profile:${_profile}, version:${_version}},{ \$set : { labels : { \"1\" : { \"name\" : "datalayer patch\", \"color\" : \"regal-rose\" }, \"2\" : { \"name\" : "Global Functions\", \"color\" : \"melon-mambo\" }}}})";

bash scripting - variable substitution and double quoting within a function

1 Answers1