从Scraped URL中抓取信息(Scrape information from Scraped URL)
我是scrapy的新手,目前正在学习如何从抓取的URL列表中抓取信息。 我已经能够通过scrapy网站上的教程从网址中抓取信息。 但是,即使在网上搜索解决方案后,我也面临从网址中抓取的网址列表中的问题抓取信息。
我在下面写的刮刀能够刮掉第一个网址。 但是,从抓取的URL列表中删除它是不成功的。 问题始于def parse_following_urls(self,response):我无法从刮掉的URL列表中删除
谁能帮忙解决这个问题? 预先感谢。
import scrapy from scrapy.http import Request class SET(scrapy.Item): title = scrapy.Field() open = scrapy.Field() hi = scrapy.Field() lo = scrapy.Field() last = scrapy.Field() bid = scrapy.Field() ask = scrapy.Field() vol = scrapy.Field() exp = scrapy.Field() exrat = scrapy.Field() exdat = scrapy.Field() class ThaiSpider(scrapy.Spider): name = "warrant" allowed_domains = ["marketdata.set.or.th"] start_urls = ["http://marketdata.set.or.th/mkt/stocklistbytype.do?market=SET&language=en&country=US&type=W"] def parse(self, response): for sel in response.xpath('//table[@class]/tbody/tr'): item = SET() item['title'] = sel.xpath('td[1]/a[contains(@href,"ssoPageId")]/text()').extract() item['open'] = sel.xpath('td[3]/text()').extract() item['hi'] = sel.xpath('td[4]/text()').extract() item['lo'] = sel.xpath('td[5]/text()').extract() item['last'] = sel.xpath('td[6]/text()').extract() item['bid'] = sel.xpath('td[9]/text()').extract() item['ask'] = sel.xpath('td[10]/text()').extract() item['vol'] = sel.xpath('td[11]/text()').extract() yield item urll = response.xpath('//table[@class]/tbody/tr/td[1]/a[contains(@href,"ssoPageId")]/@href').extract() urls = ["http://marketdata.set.or.th/mkt/"+ i for i in urll] for url in urls: request = scrapy.Request(url, callback=self.parse_following_urls, dont_filter=True) yield request request.meta['item'] = item def parse_following_urls(self, response): for sel in response.xpath('//table[3]/tbody'): item = response.meta['item'] item['exp'] = sel.xpath('tr[1]/td[2]/text()').extract() item['exrat'] = sel.xpath('tr[2]/td[2]/text()').extract() item['exdat'] = sel.xpath('tr[3]/td[2]/text()').extract() yield item
在尝试给出建议并查看输出后,我重新编写了代码。 以下是编辑过的代码。 但是,我收到另一个错误,指出
Request url must be str or unicode, got %s:' % type(url).__name__)
。 如何将URL从列表转换为字符串?我认为URL应该是字符串,因为它在For循环中。 我在下面的代码中添加了这个注释。 有什么方法可以解决这个问题吗?
import scrapy from scrapy.http import Request class SET(scrapy.Item): title = scrapy.Field() open = scrapy.Field() hi = scrapy.Field() lo = scrapy.Field() last = scrapy.Field() bid = scrapy.Field() ask = scrapy.Field() vol = scrapy.Field() exp = scrapy.Field() exrat = scrapy.Field() exdat = scrapy.Field() class ThaiSpider(scrapy.Spider): name = "warrant" allowed_domains = ["marketdata.set.or.th"] start_urls = ["http://marketdata.set.or.th/mkt/stocklistbytype.do?market=SET&language=en&country=US&type=W"] def parse(self, response): for sel in response.xpath('//table[@class]/tbody/tr'): item = SET() item['title'] = sel.xpath('td[1]/a[contains(@href,"ssoPageId")]/text()').extract() item['open'] = sel.xpath('td[3]/text()').extract() item['hi'] = sel.xpath('td[4]/text()').extract() item['lo'] = sel.xpath('td[5]/text()').extract() item['last'] = sel.xpath('td[6]/text()').extract() item['bid'] = sel.xpath('td[9]/text()').extract() item['ask'] = sel.xpath('td[10]/text()').extract() item['vol'] = sel.xpath('td[11]/text()').extract() url = ["http://marketdata.set.or.th/mkt/"]+ sel.xpath('td[1]/a[contains(@href,"ssoPageId")]/@href').extract() request = scrapy.Request(url, callback=self.parse_following_urls, dont_filter=True) #Request url must be str or unicode, got list: How to solve this? request.meta['item'] = item yield item yield request def parse_following_urls(self, response): for sel in response.xpath('//table[3]/tbody'): item = response.meta['item'] item['exp'] = sel.xpath('tr[1]/td[2]/text()').extract() item['exrat'] = sel.xpath('tr[2]/td[2]/text()').extract() item['exdat'] = sel.xpath('tr[3]/td[2]/text()').extract() yield item
I am new to scrapy and is currently learning how to scrape information from a list of scraped URL. I have been able to scrape information from a url by going thru the tutorial in scrapy website. However, i am facing problem scraping information from a list of url scraped from a url even after googling for solution online.
The scraper that i have written below is able to scrape from the first url. However, it is unsuccessful in scraping from a list of scraped URL. The problem starts at def parse_following_urls(self, response): whereby i am unable to scrape from the list of scraped URL
Can anyone help to solve this? Thank in advance.
import scrapy from scrapy.http import Request class SET(scrapy.Item): title = scrapy.Field() open = scrapy.Field() hi = scrapy.Field() lo = scrapy.Field() last = scrapy.Field() bid = scrapy.Field() ask = scrapy.Field() vol = scrapy.Field() exp = scrapy.Field() exrat = scrapy.Field() exdat = scrapy.Field() class ThaiSpider(scrapy.Spider): name = "warrant" allowed_domains = ["marketdata.set.or.th"] start_urls = ["http://marketdata.set.or.th/mkt/stocklistbytype.do?market=SET&language=en&country=US&type=W"] def parse(self, response): for sel in response.xpath('//table[@class]/tbody/tr'): item = SET() item['title'] = sel.xpath('td[1]/a[contains(@href,"ssoPageId")]/text()').extract() item['open'] = sel.xpath('td[3]/text()').extract() item['hi'] = sel.xpath('td[4]/text()').extract() item['lo'] = sel.xpath('td[5]/text()').extract() item['last'] = sel.xpath('td[6]/text()').extract() item['bid'] = sel.xpath('td[9]/text()').extract() item['ask'] = sel.xpath('td[10]/text()').extract() item['vol'] = sel.xpath('td[11]/text()').extract() yield item urll = response.xpath('//table[@class]/tbody/tr/td[1]/a[contains(@href,"ssoPageId")]/@href').extract() urls = ["http://marketdata.set.or.th/mkt/"+ i for i in urll] for url in urls: request = scrapy.Request(url, callback=self.parse_following_urls, dont_filter=True) yield request request.meta['item'] = item def parse_following_urls(self, response): for sel in response.xpath('//table[3]/tbody'): item = response.meta['item'] item['exp'] = sel.xpath('tr[1]/td[2]/text()').extract() item['exrat'] = sel.xpath('tr[2]/td[2]/text()').extract() item['exdat'] = sel.xpath('tr[3]/td[2]/text()').extract() yield item
I have re wrote the code after trying suggestions given and looking at the output. Below is the edited code. However, i got another error that states that
Request url must be str or unicode, got %s:' % type(url).__name__)
. How do i convert the URL from list to a string?I thought URL should be in string as it is in a For loop. I have added this as comment in the code below. Is there any way to solve this?
import scrapy from scrapy.http import Request class SET(scrapy.Item): title = scrapy.Field() open = scrapy.Field() hi = scrapy.Field() lo = scrapy.Field() last = scrapy.Field() bid = scrapy.Field() ask = scrapy.Field() vol = scrapy.Field() exp = scrapy.Field() exrat = scrapy.Field() exdat = scrapy.Field() class ThaiSpider(scrapy.Spider): name = "warrant" allowed_domains = ["marketdata.set.or.th"] start_urls = ["http://marketdata.set.or.th/mkt/stocklistbytype.do?market=SET&language=en&country=US&type=W"] def parse(self, response): for sel in response.xpath('//table[@class]/tbody/tr'): item = SET() item['title'] = sel.xpath('td[1]/a[contains(@href,"ssoPageId")]/text()').extract() item['open'] = sel.xpath('td[3]/text()').extract() item['hi'] = sel.xpath('td[4]/text()').extract() item['lo'] = sel.xpath('td[5]/text()').extract() item['last'] = sel.xpath('td[6]/text()').extract() item['bid'] = sel.xpath('td[9]/text()').extract() item['ask'] = sel.xpath('td[10]/text()').extract() item['vol'] = sel.xpath('td[11]/text()').extract() url = ["http://marketdata.set.or.th/mkt/"]+ sel.xpath('td[1]/a[contains(@href,"ssoPageId")]/@href').extract() request = scrapy.Request(url, callback=self.parse_following_urls, dont_filter=True) #Request url must be str or unicode, got list: How to solve this? request.meta['item'] = item yield item yield request def parse_following_urls(self, response): for sel in response.xpath('//table[3]/tbody'): item = response.meta['item'] item['exp'] = sel.xpath('tr[1]/td[2]/text()').extract() item['exrat'] = sel.xpath('tr[2]/td[2]/text()').extract() item['exdat'] = sel.xpath('tr[3]/td[2]/text()').extract() yield item
原文:https://stackoverflow.com/questions/35358867
最满意答案
每当在c ++中使用模板时,必须将使用过的模板类型称为完整类型,当您要使用字符串向量时,需要包含字符串类。 包含只不过是将包含文件中的代码复制并粘贴到您的包含所在的位置。
1> #include <string> 2> #include <vector> 3> 4> class Foo { 5> private: 6> vector<string> bar; 7> }
编译第6行时,编译器必须知道两种类型都是完整类型(字符串因为它是模板,因为它不是指针而是矢量)。 包含在类上,因此编译器在编译第6行时会知道这两种类型。无论您将它们包含在哪个顺序中都无关紧要。
Whenever you use a template in c++, the used template type has to be known as a complete type, that requires you to include the string class when you want to use a vector of string. The includes are nothing more than copying and pasting the code in the included file to where your include is placed.
1> #include <string> 2> #include <vector> 3> 4> class Foo { 5> private: 6> vector<string> bar; 7> }
When the line 6 is compiled, the compiler has to know both types as complete type (string because its a template, vector because its not a pointer). The includes are placed over the class so the compiler knows both types when he wants to compile line 6. It doesn't matter which order you included them.
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