Adding a wait-for-element while performing a SplashRequest in python Scrapy
Yes, you can write a Lua script to do that. Something like that:
function main(splash)
splash:set_user_agent(splash.args.ua)
assert(splash:go(splash.args.url))
-- requires Splash 2.3
while not splash:select('.my-element') do
splash:wait(0.1)
end
return {html=splash:html()}
end
Before Splash 2.3 you can use splash:evaljs('!document.querySelector(".my-element")')
instead of not splash:select('.my-element')
.
Save this script to a variable (lua_script = """ ... """
). Then you can send a request like this:
yield SplashRequest(
url,
self.parse,
endpoint='execute',
args={
'lua_source': lua_script,
'ua': "Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/51.0.2704.106 Safari/537.36"
}
}
See scripting tutorial and reference for more details on how to write Splash Lua scripts.
I have a similar requirement, with timeouts. My solution is a slight modification of above:
function wait_css(splash, css, maxwait)
if maxwait == nil then
maxwait = 10 --default maxwait if not given
end
local i=0
while not splash:select(css) do
if i==maxwait then
break --times out at maxwait secs
end
i=i+1
splash:wait(1) --each loop has duration 1sec
end
end