Scrapy-splash - does splash:go(url) in lua_script

Question

I'm new to Scrapy-splash and I'm trying to scrape a lazy datatable which is a table with AJAX pagination.

So I need to load the website, wait until JS is executed, get html of the table and then click on the "Next" button on pagination.

My approach works but I'm afraid I'm requesting the website two times.

First time when I yield the SplashRequest and then when lua_script is executed.

Is it true? If yes, how to make it perform request just once?

class JSSpider(scrapy.Spider):
    name = 'js_spider'
    script = """
    function main(splash, args)
        splash:go(args.url)
        splash:wait(0.5)
        local page_one = splash:evaljs("$('#example').html()")
        splash:evaljs("$('#example_next').click()")
        splash:wait(2)
        local page_two = splash:evaljs("$('#example').html()")
        return {page_one=page_one,page_two=page_two}
    end"""

    def start_requests(self):
        url = f"""https://datatables.net/examples/server_side/defer_loading.html"""
        yield SplashRequest(url, endpoint='execute',callback=self.parse, args={'wait': 0.5,'lua_source':self.script,'url':url})

    def parse(self, response):
        # assert isinstance(response, SplashTextResponse)
        page_one = response.data.get('page_one',None)
        page_one_root = etree.fromstring(page_one, HTMLParser())
        page_two = response.data.get('page_two',None)
        page_two_root = etree.fromstring(page_one, HTMLParser())

EDIT

Also I would like to wait until AJAX is performed better way than just splash:wait(2). Is it possible to somehow wait until the table changed? Ideally with some timeout.

Answer 1

Lua script is very literal - if you have 1 splash:go then one request is made by 1 splash worker.
Your crawler is fine here.

To pointlessly nit pick though: your spider connects to a worker via http so in theory two requests are being made: 1st to splash service and 2nd to target by splash worker.