nginx return 200 if rewrite_by_lua_block response header contains pattern

Question

I need nginx to return 200 code on some location only if rewrite_by_lua_block response contains custom header with pattern (for google compute engine balancer healthcheck).

I'm very newbie in lua, so any help's greatly appreciated.

Alexander Altshuler · Accepted Answer · 2016-04-13 08:04:57Z

2

I suggest to try (untested)

header_filter_by_lua_block {
   if ngx.status == 200 then
      local from, to, err = ngx.re.find( ngx.header.Foo, "your_regexp_here")
      if not from then
         ngx.status = 500 -- use the error code you need
      end
   end
}

Keep in mind - your should place header_filter_by_lua_block directive into location where your request is really sent to upstream, say where you redirect to by rewrite_by_lua_block.

answered Apr 13, 2016 at 8:04

Alexander Altshuler

3,0751 gold badge21 silver badges30 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

nginx return 200 if rewrite_by_lua_block response header contains pattern

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related