How to replace an entire NGINX block with sed?

Question

I am starting to learn bash script and need to write a script to install fastcgi_cache on NGINX. I need to replace the default location for PHP with the one with the fastcgi_cache settings.

To be clear, I need to replace this:

location ~ \.php$ {
    fastcgi_pass unix:/var/run/php5-fpm.sock;
    fastcgi_param SCRIPT_FILENAME $document_root$fastcgi_script_name;
    include fastcgi_params;
}

With this:

location ~ .php$ {
    fastcgi_pass   unix:/var/run/php5-fpm.sock ;
    fastcgi_param  SCRIPT_FILENAME  $document_root$fastcgi_script_name;
    include        fastcgi_params;

    access_log     /var/log/nginx/$SITE_URL.cache.log cache;
    fastcgi_cache_key "$mobile$scheme$request_method$host$request_uri";
    fastcgi_cache_bypass $skip_cache;
    fastcgi_no_cache $skip_cache;
    fastcgi_cache WORDPRESS;
    add_header X-Cache-Status $upstream_cache_status ;
}

I believe this can be done with sed, but I cannot find a way to do it. I managed to make it work with a sort of workaround, like this:

if [ -f "/etc/nginx/sites-enabled/$SITE_URL" ]
    then
        sed -i.bak '/fastcgi_params;/c\
        include        fastcgi_params;\
        \
        access_log     /var/log/nginx/'$SITE_URL'.cache.log cache;\
        fastcgi_cache_key "$mobile$scheme$request_method$host$request_uri";\
        fastcgi_cache_bypass $skip_cache;\
        fastcgi_no_cache $skip_cache;\
        fastcgi_cache WORDPRESS;\
        add_header X-Cache-Status $upstream_cache_status;' /etc/nginx/sites-enabled/$SITE_URL
    else
        echo ""
        echo "This domain do not exist on this server."
        echo ""
        exit 1
fi

But if the fastcgi_params pattern shows up in any other part of the file, and it might at times, it would break everything.

I am guessing that looking for the entire block, which is default and should not change across domains, is a better solution. Just cannot figure out how to do it.

The way you want to do it seems fragile. IMO this is a job for a Configuration Management tool like Ansible / Puppet. — Fredi, Oct 21 '16 at 16:37
@Fredi Though about that, but our main system admin does not want to do it and he has the keys for Puppet. I have the documentation for my team but figured I would write a bash script to make it faster for them. — James Morningstar, Oct 21 '16 at 17:12
The whole point of Puppet is so that you don't have to do ridiculous time-wasting stuff like this. Push back harder. — Michael Hampton, Dec 17 '17 at 19:00

score 1 · Answer 1 · edited Aug 22 '19 at 13:40

This should work for you.

sed -i '/location.*php/{:a;N;/fastcgi_pass/{N;N;d};/  }/b;ba}' <filename>
cat >> <filename> << EOF
location ~ .php$ {
    fastcgi_pass   unix:/var/run/php5-fpm.sock ;
    fastcgi_param  SCRIPT_FILENAME  $document_root$fastcgi_script_name;
    include        fastcgi_params;
    access_log     /var/log/nginx/$SITE_URL.cache.log cache;
    fastcgi_cache_key "$mobile$scheme$request_method$host$request_uri";
    fastcgi_cache_bypass $skip_cache;
    fastcgi_no_cache $skip_cache;
    fastcgi_cache WORDPRESS;
    add_header X-Cache-Status $upstream_cache_status ;
}
EOF

As mentioned in the above posts... This is not a best-practice solution compared to let's say a configuration manager such as Puppet, Ansible or Salt.

The outcomes of running these operations on a broken nginx.conf, would be unpredictable.

How to replace an entire NGINX block with sed?

1 Answers1