To remove line based on string - shell

Question

I have file like test.yaml file, the text content in the file like below.
servers:
- uri: "http://demo.nginx1.com/status"
name: "NGinX Monitor1"
- uri: "http://demo.nginx2.com/status"
name: "NGinX Monitor2"
I want to remove - uri line and immediate next line (start with name:) where host name = demo.nginx1.com.
I want out put like below.
servers:
- uri: "http://demo.nginx2.com/status"
name: "NGinX Monitor2"
I tied like below..
cat test.yaml | grep -v demo.nginx1.com | grep -v Monitor1 >> test_back.yaml
mv test_back.yaml test.yaml
I am getting expected out put. But it's re creating the file and I don't want to re create the file
Please help me with suitable command that i can use..

Answer 1

Just a simple logic using GNU sed
sed '/demo.nginx1.com/,+1 d' test.yaml
servers:
- uri: "http://demo.nginx2.com/status"
name: "NGinX Monitor2"
For in-place replacement, add a -i flag as -i.bak
sed -i.bak '/demo.nginx1.com/,+1 d' test.yaml
To see the in-place replacement:-
cat test.yaml
servers:
- uri: "http://demo.nginx2.com/status"
name: "NGinX Monitor2"

Answer 2

As I dislike using regular expressions to hack something you can parse - here's how I'd tackle it, using perl and the YAML module:
#!/usr/bin/env perl
use strict;
use warnings;
use YAML;
use Data::Dumper;
#load yaml by reading files specified as args to stdin,
#or piped in. (Just like how you'd use 'sed')
my $config = Load ( do { local $/ ; <>} );
#output data structure for debug
print Dumper $config;
#use grep to filter the uri you don't want.
#{$config -> {servers}} = grep { not $_ -> {uri} =~ m/demo.nginx2/ } #{$config -> {servers}};
#resultant data structure
print Dumper $config;
#output YAML to STDOUT
print Dump $config;