Bash script double quote escaping

Question

I'm trying to run a command in a bash script using a variable as a parameter, but I am failing to correctly escape the quotes. This is what I have have:

vsnames=$(comm -13 --nocheck-order $cur_ha_stat_report $old_ha_stat_report | awk -F\| '{print $1}' | tr '\r\n' ' ')
create event type=0xfff00033 text="$vsnames" mh=$dev_mh &

Where vsnames is a string with words separated by spaces. When I run the script, this is what I get:

+ vsnames='VS1 VS2 '
+ create event type=0xfff00033 'text=VS1 VS2 ' mh=xxx

How to I put the $vsnames variable between quotes? I have tried using \\" but it didn't work, bash just added a bunch of unwanted single-quotes. Any tips? The command I actually need is: create event type=0xfff00033 text="VS1 VS2 " mh=xxx

Answer 1

With your variable vsnames being set to the 8-character-string VS1 VS2 (as we can see from the trace), doing a

create event type=0xfff00033 text=\""$vsnames"\" mh=xxx

would set the argv[2] of the create command to

text="VS1 VS2 "

Explanation: I have enclosed $vsnames into single quotes, to tell bash that we don't want to do word splitting here. Then I wrapped this between literal \\" quotes, because you want create to see these quotes.