python - Scrapy跨多个表单请求页面存储项目？元？ Python

Question

所以我让我的抓取工具可以处理一个表单请求。我什至可以看到终端打印出单页版本中的抓取数据:

class MySpider(BaseSpider):
    name = "swim"
    start_urls = ["example.website"]
    DOWNLAD_DELAY= 30.0

    def parse(self, response):
        return [FormRequest.from_response(response,formname="TTForm",
                    formdata={"Ctype":"A", "Req_Team": "", "AgeGrp": "0-6", 
                    "lowage": "", "highage": "", "sex": "W", "StrkDist": "10025", 
                    "How_Many": "50", "foolOldPerl": ""}
                    ,callback=self.swimparse1,dont_click=True)]

    def swimparse1(self, response):       
        open_in_browser(response)
        hxs = Selector(response)
        rows = hxs.xpath(".//tr")
        items = []

        for rows in rows[4:54]:
            item = swimItem()
            item["names"] = rows.xpath(".//td[2]/text()").extract()
            item["age"] = rows.xpath(".//td[3]/text()").extract()
            item["free"] = rows.xpath(".//td[4]/text()").extract()
            item["team"] = rows.xpath(".//td[6]/text()").extract()
            items.append(item)

        return items

但是，当我添加第二个表单请求回调时，它只会抓取第二个表单中的项目。它还只打印第二页的抓取，就好像它完全跳过第一页抓取一样？ :

class MySpider(BaseSpider):
    name = "swim"
    start_urls = ["example.website"]
    DOWNLAD_DELAY= 30.0

    def parse(self, response):
        return [FormRequest.from_response(response,formname="TTForm",
                    formdata={"Ctype":"A", "Req_Team": "", "AgeGrp": "0-6", 
                    "lowage": "", "highage": "", "sex": "W", "StrkDist": "10025", 
                    "How_Many": "50", "foolOldPerl": ""}
                    ,callback=self.swimparse1,dont_click=True)]

    def swimparse1(self, response):       
        open_in_browser(response)
        hxs = Selector(response)
        rows = hxs.xpath(".//tr")
        items = []

        for rows in rows[4:54]:
            item = swimItem()
            item["names"] = rows.xpath(".//td[2]/text()").extract()
            item["age"] = rows.xpath(".//td[3]/text()").extract()
            item["free"] = rows.xpath(".//td[4]/text()").extract()
            item["team"] = rows.xpath(".//td[6]/text()").extract()
            items.append(item)
            #print item[]           
        return [FormRequest.from_response(response,formname="TTForm",
                    formdata={"Ctype":"A", "Req_Team": "", "AgeGrp": "0-6", 
                    "lowage": "", "highage": "", "sex": "W", "StrkDist": "40025", 
                    "How_Many": "50", "foolOldPerl": ""}
                    ,callback=self.Swimparse2,dont_click=True),]

    def swimparse2(self, response):
        open_in_browser(response)
        hxs = Selector(response)
        rows = hxs.xpath(".//tr")
        items = []

        for rows in rows[4:54]:
            item = swimItem()
            item["names"] = rows.xpath(".//td[2]/text()").extract()
            item["age"] = rows.xpath(".//td[3]/text()").extract()
            item["fly"] = rows.xpath(".//td[4]/text()").extract()
            item["team"] = rows.xpath(".//td[6]/text()").extract()
            items.append(item)
            #print item[]
        return items

猜测:A)如何将第一次抓取中的项目导出或返回到第二次抓取中，以便最终将所有项目数据放在一起，就像从一页抓取一样？

B)或者如果第一个抓取被完全跳过，我怎样才能停止跳过并将这些项目传递到下一个？

谢谢!

PS:另外:我尝试过使用:

item = response.request.meta = ["item]
item = response.request.meta = []
item = response.request.meta = ["names":item, "age":item, "free":item, "team":item]

所有这些都会引发关键错误或其他异常

我还尝试修改表单请求以包含 meta={"names":item, "age":item, "free":item, "team":item}。不会引发错误，但不会抓取或存储任何内容。

编辑:我尝试使用这样的产量:

class MySpider(BaseSpider):
name = "swim"
start_urls = ["www.website.com"]
DOWNLAD_DELAY= 30.0

def parse(self, response):
    open_in_browser(response)
    hxs = Selector(response)
    rows = hxs.xpath(".//tr")
    items = []

    for rows in rows[4:54]:
        item = swimItem()
        item["names"] = rows.xpath(".//td[2]/text()").extract()
        item["age"] = rows.xpath(".//td[3]/text()").extract()
        item["free"] = rows.xpath(".//td[4]/text()").extract()
        item["team"] = rows.xpath(".//td[6]/text()").extract()
        items.append(item) 
        yield [FormRequest.from_response(response,formname="TTForm",
                formdata={"Ctype":"A", "Req_Team": "", "AgeGrp": "0-6", 
                "lowage": "", "highage": "", "sex": "W", "StrkDist": "10025", 
                "How_Many": "50", "foolOldPerl": ""}
                ,callback=self.parse,dont_click=True)]

    for rows in rows[4:54]:
        item = swimItem()
        item["names"] = rows.xpath(".//td[2]/text()").extract()
        item["age"] = rows.xpath(".//td[3]/text()").extract()
        item["fly"] = rows.xpath(".//td[4]/text()").extract()
        item["team"] = rows.xpath(".//td[6]/text()").extract()
        items.append(item)

        yield [FormRequest.from_response(response,formname="TTForm",
                formdata={"Ctype":"A", "Req_Team": "", "AgeGrp": "0-6", 
                "lowage": "", "highage": "", "sex": "W", "StrkDist": "40025", 
                "How_Many": "50", "foolOldPerl": ""}
                ,callback=self.parse,dont_click=True)]

仍然没有抓取任何东西。我知道 xpath 是正确的，因为当我只尝试并抓取一种形式(使用返回而不是 yield )时，它工作得很好。我已经阅读了杂乱的文档，但它并不是很有帮助:(

Answer 1

您缺少一个非常简单的解决方案，请将 return 更改为 yield

然后，您不必在数组中累积项目，只需从函数中生成所需数量的项目和请求，scrapy 将完成剩下的工作

来自scrapy docs :

from scrapy.selector import Selector
from scrapy.spider import Spider
from scrapy.http import Request
from myproject.items import MyItem

class MySpider(Spider):
    name = 'example.com'
    allowed_domains = ['example.com']
    start_urls = [
        'http://www.example.com/1.html',
        'http://www.example.com/2.html',
        'http://www.example.com/3.html',
    ]

    def parse(self, response):
        sel = Selector(response)
        for h3 in sel.xpath('//h3').extract():
            yield MyItem(title=h3)

        for url in sel.xpath('//a/@href').extract():
            yield Request(url, callback=self.parse)