如何使用python在scrapy中迭代XML子节点？

我想在上面刮注释，但我似乎不知道如何遍历封装注释的节点的子节点并获取数据点

这是hmtl的一部分：

#跟在后面
我认为您的问题来自“注释”的xpath。通过仅获取文本，您没有选择节点。
以下更改使其适用于我：

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                    <a href="https://www.picuki.com/profile/alexandera_300">
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                    <a href="https://www.picuki.com/profile/alexandera_300">@alexandera_300</a>
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                #followforfollowback
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                    <a href="https://www.picuki.com/profile/coxlogan2008">@coxlogan2008</a>
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            <div class="comment-text">
                I think your issue comes from your xpath for 'comments'. By taking only the text, you're not selecting the nodes.
The following changes make it work for me:


# the likes & number of comments only have to be taken once, should not be part of the loop
likes = response.xpath('.//span[@class="icon-thumbs-up-alt"]/text()').get()
num_of_comments = response.xpath('.//span[@id="commentsCount"]/text()').get()
comments = response.xpath('//div[@id="commantsPlace"]/*[@class="comment"]')
for comment in comments:  
    comment_user_name = comment.xpath('.//*[@class="comment-user-nickname"]/a/text()').get()
    comment_text = comment.xpath('.//*[@class="comment-text"]/text()').get()