Splash Lua 脚本附近预期错误答案

【问题标题】：Error expected near on Splash Lua ScriptSplash Lua 脚本附近预期错误
【发布时间】：2017-07-14 15:48:20
【问题描述】：

我需要在 Splash 上运行这个简单的 Lua 脚本：

    script = """
    function main(splash)
        local url = splash.args.url
        assert(splash:go(url))
        assert(splash:wait(1))

        -- go back 1 month in time and wait a little (1 second)
        assert(splash:runjs("$('a[data-ci-test-id=\"checkOutButton\"]').click()"))
        assert(splash:wait(1))

        -- return result as a JSON object
        return {
            html = splash:html(),
            -- we don't need screenshot or network activity
            --png = splash:png(),
            --har = splash:har(),
        }
    end
    """

但是我收到了这个错误：

{u'info': {u'source': u'[string "..."]', u'message': u'[string "..."]:8: \')\' expected near \'checkOutButton\'', u'line_number': 8, u'type': u'LUA_INIT_ERROR', u'error': u"')' expected near 'checkOutButton'"}, u'type': u'ScriptError', u'description': u'Error happened while executing Lua script', u'error': 400}

我应该如何编写 CSS 选择器字符串才能使该脚本正常工作？

【问题讨论】：

标签： python lua syntax-error scrapy-splash splash-js-render

【解决方案1】：

即使错误字符串不相同，我也根据另一个答案找到了答案。 Use scrapy + splash return html

原来，我抓取的页面没有JQuery，所以我必须先加载JQuery包。

    script = """
    function main(splash)
        # add this vvv
        assert(splash:autoload("https://code.jquery.com/jquery-3.1.1.min.js"))
        local url = splash.args.url
        assert(splash:go(url))
        assert(splash:wait(1))

        -- go back 1 month in time and wait a little (1 second)
        assert(splash:runjs("$('div.minicart_summery > a.btn_checkout').click()"))
        assert(splash:wait(1))

        -- return result as a JSON object
        return {
            html = splash:html(),
            -- we don't need screenshot or network activity
            --png = splash:png(),
            --har = splash:har(),
        }
    end
    """

【讨论】：