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我正在尝试解析网站的多个页面，但我不明白如何更改 url 的查询（如果这有意义？）

我尝试创建一个 next_page ，它每次找到下一页元素时都会添加第一页并添加 +1，但我认为我不能，因为我将有多个起始 url（都相似）。当我尝试获取下一页元素的信息时，它返回：

["loadmoreresult('?networkId=24&pageNumber=2&pageSize=100&allnet=yes&networkIds=1&networkIds=2&networkIds=3&networkIds=4&networkIds=61&networkIds=98&networkIds=108&networkIds=6&networkIds=5&networkIds=22&networkIds=13&networkIds=18&networkIds=15&networkIds=16&networkIds=105&networkIds=38&licenseIds=0&licenseIds= 0&licenseIds=0&licenseIds=0&licenseIds=0&searchby=CountryCode&orderby=CountryCity&country=ES&city=&keyword=&lastCid=116490');返回false;"]

使用 url.parse(response.url).query 我得到：

'networkId=24&pageNumber=1&pageSize=100&allnet=yes&networkIds=1&networkIds=2&networkIds=3&networkIds=4&networkIds=61&networkIds=98&networkIds=108&networkIds=6&networkIds=5&networkIds=22&networkIds=13&networkIds=18&networkIds=15&networkIds=16&networkIds=105&networkIds=38&licenseIds=0&licenseIds=0&licenseIds=0&licenseIds=0&licenseIds =0&searchby=CountryCode&orderby=CountryCity&country=ES&city=&keyword='

我需要做的就是创建一个使用相同方案、路径的新链接，然后更改查询。

如果您需要更多信息，请告诉我，我真的不知道什么与您更相关，因为我还是一个初学者。

from urllib.parse import urlparse, urljoin

urlparse(response.url)
>>> ParseResult(scheme='https', netloc='www.wcaworld.com', path='/Directory', params='', query='networkId=24&pageNumber=1&pageSize=100&allnet=yes&networkIds=1&networkIds=2&networkIds=3&networkIds=4&networkIds=61&networkIds=98&networkIds=108&networkIds=6&networkIds=5&networkIds=22&networkIds=13&networkIds=18&networkIds=15&networkIds=16&networkIds=105&networkIds=38&licenseIds=0&licenseIds=0&licenseIds=0&licenseIds=0&licenseIds=0&searchby=CountryCode&orderby=CountryCity&country=ES&city=&keyword=', fragment='')

response.css('a.loadmore::attr(onmouseover)').extract()
>>>["loadmoreresult('?networkId=24&pageNumber=2&pageSize=100&allnet=yes&networkIds=1&networkIds=2&networkIds=3&networkIds=4&networkIds=61&networkIds=98&networkIds=108&networkIds=6&networkIds=5&networkIds=22&networkIds=13&networkIds=18&networkIds=15&networkIds=16&networkIds=105&networkIds=38&licenseIds=0&licenseIds=0&licenseIds=0&licenseIds=0&licenseIds=0&searchby=CountryCode&orderby=CountryCity&country=ES&city=&keyword=&lastCid=116490'); return false;"]
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1 回答 1

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您需要获取该<a>元素的基本 url，它是查询字符串开始之前 url 的一部分，https://example.com/a/path/?query=param所以这里的基本 url 将是https://example.com/a/path/. 将其保存到变量中。然后用于urllib.parse.parse_qsl解析查询字符串，然后更新页码并将其与基本 url 连接。

from urllib.parse import parse_qsl, urljoin, urlencode

BASE_URL = 'https://example.com/a/path/'
# you can also extract base url from scrapy.Response object
# BASE_URL, _ = splitquery(response.url)

if __name__ == '__main__':
    # extract query parameter from from a url
    q = 'networkId=24&pageNumber=2&pageSize=100&allnet=yes&networkIds=1&networkIds=2&networkIds=3&networkIds=4&networkIds=61&networkIds=98&networkIds=108&networkIds=6&networkIds=5&networkIds=22&networkIds=13&networkIds=18&networkIds=15&networkIds=16&networkIds=105&networkIds=38&licenseIds=0&licenseIds=0&licenseIds=0&licenseIds=0&licenseIds=0&searchby=CountryCode&orderby=CountryCity&country=ES&city=&keyword=&lastCid=116490'
    parsed = dict(parse_qsl(q))
    next_page = int(parsed['pageNumber']) + 1
    parsed['pageNumber'] = next_page

    next_page_url = urljoin(BASE_URL, '?' + urlencode(parsed))

    print(next_page_url)

输出：

https://example.com/a/path/networkId=24&pageNumber=3&pageSize=100&allnet=yes&networkIds=38&licenseIds=0&searchby=CountryCode&orderby=CountryCity&country=ES&lastCid=116490
于 2019-07-16T22:10:18.997 回答