AWK print block that does NOT contain specific text - bash

Question

I have the following data file:
variable "ARM_CLIENT_ID" {
description = "Client ID for Service Principal"
}
variable "ARM_CLIENT_SECRET" {
description = "Client Secret for Service Principal"
}
# [.....loads of code]
variable "logging_settings" {
description = "Logging settings from TFVARs"
}
variable "azure_firewall_nat_rule_collections" {
default = {}
}
variable "azure_firewall_network_rule_collections" {
default = {}
}
variable "azure_firewall_application_rule_collections" {
default = {}
}
variable "build_route_tables" {
description = "List of Route Table keys that need direct internet prior to Egress FW build"
default = [
"shared_services",
"sub_to_afw"
]
}
There's a 2 things I wish to do:
print the variable names without the inverted commas
ONLY print the variables names if the code block does NOT contain default
I know I can print the variables names like so: awk '{ gsub("\"", "") }; (/variable/ && $2 !~ /^ARM_/) { print $2}'
I know I can print the code blocks with: awk '/variable/,/^}/', which results:
# [.....loads of code output before this]
variable "logging_settings" {
description = "Logging settings from TFVARs"
}
variable "azure_firewall_nat_rule_collections" {
default = {}
}
variable "azure_firewall_network_rule_collections" {
default = {}
}
variable "azure_firewall_application_rule_collections" {
default = {}
}
variable "build_route_tables" {
description = "List of Route Table keys that need direct internet prior to Egress FW build"
default = [
"shared_services",
"sub_to_afw"
]
}
However, I cannot find out how to print the code blocks "if" they don't contain default. I know I will need to use an if statement, and some variables perhaps, but I am unsure as of how.
This code block should NOT appear in the output for which I grab the variable name:
variable "build_route_tables" {
description = "List of Route Table keys that need direct internet prior to Egress FW build"
default = [
"shared_services",
"sub_to_afw"
]
}
End output should NOT contain those that had default:
# [.....loads of code output before this]
expressroute_settings
firewall_settings
global_settings
peering_settings
vnet_transit_object
vnet_shared_services_object
route_tables
logging_settings
Preferable I would like to keep this a single AWK command or file, no piping. I have uses for this that do prefer no piping.
EDIT: update the ideal outputs (missed some examples of those with default)

Answer 1

Assumptions and collection of notes from OP's question and comments:
all variable definition blocks end with a right brace (}) in the first column of a new line
we only display variable names (sans the double quotes)
we do not display the variable names if the body of the variable definition contains the string default
we do not display the variable name if it starts with the string ARM_
One (somewhat verbose) awk solution:
NOTE: I've copied the sample input data into my local file variables.dat
awk -F'"' ' # use double quotes as the input field separator
/^variable / && $2 !~ "^ARM_" { varname = $2 # if line starts with "^variable ", and field #2 is not like "^ARM_", save field #2 for later display
printme = 1 # enable our print flag
}
/variable/,/^}/ { if ( $0 ~ "default" ) # within the range of a variable definition, if we find the string "default" ...
printme = 0 # disable the print flag
next # skip to next line
}
printme { print varname # if the print flag is enabled then print the variable name and then ...
printme = 0 # disable the print flag
}
' variables.dat
This generates:
logging_settings

Answer 2

$ awk -v RS= '!/default =/{gsub(/"/,"",$2); print $2}' file
ARM_CLIENT_ID
ARM_CLIENT_SECRET
[.....loads
logging_settings
of course output doesn't match yours since it's inconsistent with the input data.

Answer 3

Using GNU awk:
awk -v RS="}" '/variable/ && !/default/ && !/ARN/ { var=gensub(/(^.*variable ")(.*)(".*{.*)/,"\\2",$0);print var }' file
Set the record separator to "}" and then check for records that contain "variable", don't contain default and don't contain "ARM". Use gensub to split the string into three sections based on regular expressions and set the variable var to the second section. Print the var variable.
Output:
logging_settings

Answer 4

Another variation on awk using skip variable to control the array index holding the variable names:
awk '
/^[[:blank:]]*#/ { next }
$1=="variable" { gsub(/["]/,"",$2); vars[skip?n:++n]=$2; skip=0 }
$1=="default" { skip=1 }
END { if (skip) n--; for(i=1; i<=n; i++) print vars[i] }
' code
The first rule just skips comment lines. If you want to skip "ARM_" variables, then you can add a test on $2.
Example Use/Output
With your example code in code, all variables without default are:
$ awk '
> /^[[:blank:]]*#/ { next }
> $1=="variable" { gsub(/["]/,"",$2); vars[skip?n:++n]=$2; skip=0 }
> $1=="default" { skip=1 }
> END { if (skip) n--; for(i=1; i<=n; i++) print vars[i] }
> ' code
ARM_CLIENT_ID
ARM_CLIENT_SECRET
logging_settings

Answer 5

Here's another maybe shorter solution.
$ awk -F'"' '/^variable/&&$2!~/^ARM_/{v=$2} /default =/{v=0} /}/&&v{print v; v=0}' file
logging_settings