python - Scrapy 中的项目编号

Question

所以我有一个包含以下内容的 items.py:

class ScrapyItem(scrapy.Item):
    source = scrapy.Field()
    link = scrapy.Field()

json输出为:

[{"source": "Some source", "link":"www.somelink.com"},
 {"source": "Some source again", "link":"www.somelink.org"}]

有没有办法将输出更改为:

[{"source1": "Some source", "link1":"www.somelink.com"},
 {"source2": "Some source again", "link2":"www.somelink.org"}]

从文档中，我看到您可以操纵项目值，您可以对项目本身做同样的事情吗？

编辑

这是我用于输出带有 article_id item_field 的新代码

article_id = [1]
def parse_common(self, response):
    feed = feedparser.parse(response.body)
    for entry_n, entry in enumerate(feed.entries, start=article_id[-1]):
        try:
            item = NewsbyteItem()
            item['source'] = response.url
            item['title'] = lxml.html.fromstring(entry.title).text
            item['link'] = entry.link
            item['description'] = entry.description
            item['article_id'] = '%d' % entry_n
            article_id.append(entry_n)
            request = Request(
                entry.link,
                callback=getattr(self, response.meta['method']),
                dont_filter=response.meta.get('dont_filter', False)
            )

            request.meta['item'] = item
            request.meta['entry'] = entry

            yield request
        except Exception as e:
            print '%s: %s' % (type(e), e)
            print entry

问题是 entry_n 在更改为另一个 url 时会重新启动。这就是使用该列表的原因。

Answer 1

我不建议您识别不同的项目来更改项目值的键。您可以使用字典来命名响应，例如:

output = [{"source": "Some source", "link":"www.somelink.com"}, {"source": "Some source again", "link":"www.somelink.org"}]
output_dict = {}
for counter, item in enumerate(output):
    output_dict['item' + str(counter + 1)] = item
print output_dict