title: I just want to use splash script and return only job listing's urls
body: <p>I'm using scrapy and splash to scrape <a href="http://www.gabteo.ro/jobs/" rel="nofollow noreferrer">this</a> website, for some reasons I'm using splash and scrapy even though I know I can scrape its API. My problem is that I only want my lua script to return only job listing's <code>urls</code> rather than the whole <code>splash:html()</code> page, I've been trying to do that but I'm getting the error message below:-</p>
  
  <pre><code> {
      "error": 400,
      "description": "Error happened while executing Lua script",
      "type": "ScriptError",
      "info": {
          "message": "Lua error: /app/splash/lua_modules/libs/treat.lua:45: cannot change a protected metatable",
          "type": "LUA_ERROR"
      }
  }
  </code></pre>
  
  <p>The lua script I've been using is also shown below:-</p>
  
  <pre><code>function main(splash, args)
      assert(splash:go(args.url))
      splash:wait(5.0)
      local treat = require('treat')
      listings = assert(splash:select_all("ul.job_listings &gt; li&gt; a"))
  
      return {
         listing_urls = treat.as_array(listings)
      }
  
  end
  </code></pre>