Python xpath => SyntaxError: invalid syntax

Question

I am a newbie to python. I'm trying to crawl some text, but it doesn't work... Here is my code

    def parse(self,response):

    hxs = Selector(response)
    article_list = hsx.xpath('//*[@id="weather"]')

    print('##################### total_length : ' + str(len(article_list)))

    for a in article_list:
        html_response = HtmlResponse(uurl="", body=a.encode('utf-8') #??
        seoul = html_response.xpath('//dl[@class="po_seoul"]//dd[@class="temp"]')[0].extract()

and this is the html code.

            <h1 class="blind">weather_data</h1>
            <div id="weather" class="weather" >
                <dl class="po_seoul">
                    <dt>seoul</dt>
                    <dd class="weather"><a href="#" onclick="showLandForecast('1159068000'); return false;" data-stn="108"><img src="/images/icon/NW/NB04.png" alt="cloud" /></a></dd>
                    <dt class="blind">temperature</dt>
                    <dd class="temp">21.2</dd>
                </dl>

Error message:

seoul = html_response.xpath('//dl[@class="po_seoul"]//dd[@class="temp"]')[0].extract() ^ SyntaxError: invalid syntax

plz help me...T0T

html_response = HtmlResponse(uurl="", body=a.encode('utf-8') you are missing closing parentheses in this line. — Lafexlos
– Lafexlos, Commented Oct 5, 2017 at 9:10

mantkowicz · Accepted Answer · 2017-10-05 09:10:18Z

1

You are missing ) after HtmlResponse(uurl="", body=a.encode('utf-8') it should be

HtmlResponse(uurl="", body=a.encode('utf-8'))
                                            ^

instead of what you have

answered Oct 5, 2017 at 9:10

mantkowicz

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