[英]Scraping Multiple Pages Scrapy
我试图每年都爬到Billboard的前100名中。我有一个文件,可以一次使用一年,但是我希望它可以在所有年份中抓取并收集这些数据。 这是我当前的代码:
from scrapy import Spider
from scrapy.selector import Selector
from Billboard.items import BillboardItem
from scrapy.exceptions import CloseSpider
from scrapy.http import Request
URL = "http://www.billboard.com/archive/charts/%/hot-100"
class BillboardSpider(Spider):
name = 'Billboard_spider'
allowed_urls = ['http://www.billboard.com/']
start_urls = [URL % 1958]
def _init_(self):
self.page_number=1958
def parse(self, response):
print self.page_number
print "----------"
rows = response.xpath('//*[@id="block-system-main"]/div/div/div[2]/table/tbody/tr').extract()
for row in rows:
IssueDate = Selector(text=row).xpath('//td[1]/a/span/text()').extract()
Song = Selector(text=row).xpath('//td[2]/text()').extract()
Artist = Selector(text=row).xpath('//td[3]/a/text()').extract()
item = BillboardItem()
item['IssueDate'] = IssueDate
item['Song'] = Song
item['Artist'] = Artist
yield item
self.page_number += 1
yield Request(URL % self.page_number)
但出现错误:“ start_urls = [URL%1958] ValueError:索引41处不支持的格式字符'/'(0x2f)”
有任何想法吗? 我希望代码从原始的“ URL”链接自动将年份更改为1959,并逐年进行下去,直到它停止查找表格,然后关闭。
您收到的错误是因为您没有使用正确的语法进行字符串格式化。 您可以在这里查看其工作方式的详细信息。 在特定情况下不起作用的原因是您的URL缺少“ s”:
URL = "http://www.billboard.com/archive/charts/%/hot-100"
应该
URL = "http://www.billboard.com/archive/charts/%s/hot-100"
无论如何,最好使用新样式的字符串格式:
URL = "http://www.billboard.com/archive/charts/{}/hot-100"
start_urls = [URL.format(1958)]
继续,您的代码还有其他一些问题:
def _init_(self):
self.page_number=1958
如果要使用init函数,则应将其命名为__init__
(两个下划线),并且由于要扩展Spider
,因此需要传递*args
和**kwargs
以便可以调用父构造函数:
def __init__(self, *args, **kwargs):
super(MySpider, self).__init__(*args, **kwargs)
self.page_number = 1958
这听起来像你可能会关闭不使用更好的__init__
,而是只用一个列表理解生成所有从一开始走的网址:
start_urls = ["http://www.billboard.com/archive/charts/{year}/hot-100".format(year=year)
for year in range(1958, 2017)]
start_urls
将如下所示:
['http://www.billboard.com/archive/charts/1958/hot-100',
'http://www.billboard.com/archive/charts/1959/hot-100',
'http://www.billboard.com/archive/charts/1960/hot-100',
'http://www.billboard.com/archive/charts/1961/hot-100',
...
'http://www.billboard.com/archive/charts/2017/hot-100']
您还没有正确填充BillboardItem
,因为对象(默认情况下)不支持项目分配:
item = BillboardItem()
item['IssueDate'] = IssueDate
item['Song'] = Song
item['Artist'] = Artist
应该:
item = BillboardItem()
item.IssueDate = IssueDate
item.Song = Song
item.Artist = Artist
尽管通常最好在类的init函数中执行此操作:类BillboardItem(object):def init (自我,issue_date,歌曲,歌手):self.issue_date = issue_date self.song =歌曲self.artist = artist然后通过item = BillboardItem(IssueDate, Song, Artist)
创建项目
无论如何,我清理了您的代码(并创建了BillboardItem,因为我不完全了解您的外观):
from scrapy import Spider, Item, Field
from scrapy.selector import Selector
from scrapy.exceptions import CloseSpider
from scrapy.http import Request
class BillboardItem(Item):
issue_date = Field()
song = Field()
artist = Field()
class BillboardSpider(Spider):
name = 'billboard'
allowed_urls = ['http://www.billboard.com/']
start_urls = ["http://www.billboard.com/archive/charts/{year}/hot-100".format(year=year)
for year in range(1958, 2017)]
def parse(self, response):
print(response.url)
print("----------")
rows = response.xpath('//*[@id="block-system-main"]/div/div/div[2]/table/tbody/tr').extract()
for row in rows:
issue_date = Selector(text=row).xpath('//td[1]/a/span/text()').extract()
song = Selector(text=row).xpath('//td[2]/text()').extract()
artist = Selector(text=row).xpath('//td[3]/a/text()').extract()
item = BillboardItem(issue_date=issue_date, song=song, artist=artist)
yield item
希望这可以帮助。 :)
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