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重要提示：目前在 stackoverflow 上的所有答案都适用于以前版本的 Scrapy，不适用于最新版本的 scrapy 1.4

对scrapy和python完全陌生，我正在尝试抓取一些页面并下载图像。正在下载图像，但它们仍然具有原始 SHA-1 名称作为文件名。我不知道如何重命名文件，它们实际上都有 SHA-1 文件名。

试图将它们重命名为“test”，当我运行时，我确实在输出中出现了“test” scrapy crawl rambopics，以及 url 的数据。但文件不会在目标文件夹中重命名。以下是输出示例：

> 2017-06-11 00:27:06 [scrapy.core.scraper] DEBUG: Scraped from <200
> http://www.theurl.com/> {'image_urls':
> ['https://www.theurl.com/-a4Bj-ENjHOY/VyE1mGuJyUI/EAAAAAAAHMk/mw1_H-mEAc0QQEwp9UkTipxNCVR-xdbcgCLcB/s1600/Image%2B%25286%2525.jpg'],
> 'image_name': ['test'], 'title': ['test'], 'filename': ['test'],
> 'images': [{'url':
> 'https://www.theurl.com/-a4Bj-ENjHOY/VyE1mGuJyUI/EAAAAAAAHMk/mw1_H-mEAc0QQEwp9UkTipxNCVR-xdbcgCLcB/s1600/Image%2B%25286%2525.jpg',
> 'path': 'full/fcbec9bf940b48c248213abe5cd2fa1c690cb879.jpg',
> 'checksum': '7be30d939a7250cc318e6ef18a6b0981'}]}

到目前为止，我已经尝试了许多不同的解决方案，都发布在 stackoverflow 上，对于 2017 年最新版本的 scrapy，这个问题没有明确的答案，看起来这些命题可能几乎都已经过时了。我将 Scrapy 1.4 与 python 3.6 一起使用。

scrapy.cfg

# Automatically created by: scrapy startproject
#
# For more information about the [deploy] section see:
# https://scrapyd.readthedocs.org/en/latest/deploy.html

[settings]
default = rambopics.settings

[deploy]
#url = http://localhost:6800/
project = rambopics

items.py 导入scrapy

class RambopicsItem(scrapy.Item):
    # defining items:
     image_urls = scrapy.Field()
     images = scrapy.Field()
     image_name = scrapy.Field()
     title = scrapy.Field()
    #pass -- dont realy understand what pass is for

设置.py

BOT_NAME = 'rambopics'

SPIDER_MODULES = ['rambopics.spiders']
NEWSPIDER_MODULE = 'rambopics.spiders'


ROBOTSTXT_OBEY = True


ITEM_PIPELINES = {
   'scrapy.pipelines.images.ImagesPipeline': 1,
}

IMAGES_STORE = "W:/scraped/"

管道.py

import scrapy
from scrapy.pipelines.images import ImagesPipeline

class RambopicsPipeline(ImagesPipeline):


    def get_media_requests(self, item, info):

        img_url = item['img_url']
        meta = {
                  'filename': item['title'],
                   'title': item['image_name']
                }

        yield Request(url=img_url, meta=meta)

（蜘蛛）rambopics.py

from rambopics.items import RambopicsItem
from scrapy.selector import Selector
import scrapy


class RambopicsSpider(scrapy.Spider):
    name = 'rambopics'
    allowed_domains = ['theurl.com']
    start_urls = ['http://www.theurl.com/']

    def parse(self, response):

        for sel in response.xpath('/html'):
            #img_name = sel.xpath("//h3[contains(@class, 'entry-title')]/a/text()").extract()
            img_name = 'test'
            #img_title = sel.xpath("//h3[contains(@class, 'entry-title')]/a/text()").extract()
            img_title = 'test' 

        for elem in response.xpath("//div[contains(@class, 'entry-content')]"):
            img_url = elem.xpath("a/@href").extract_first()


            yield {
                   'image_urls': [img_url],
                   'image_name': [img_name],
                   'title': [img_title],
                   'filename': [img_name]
               }

请注意，我不知道最终下载的文件名使用的正确元名称是什么（我不确定它是文件名、图像名还是标题）。

4

1 回答 1

1

使用file_path方法更改图像名称，如下所示：

class SaveImagesPipeline(FilesPipeline):
    def get_media_requests(self, item, info):
        i = 1
        for image_url in item['image_urls']:
            filename = '{}_{}.jpg'.format(item['name_image'], i)
            yield scrapy.Request(image_url, meta={'filename': filename})
            i += 1
    return

    def file_path(self, request, response=None, info=None):
        return request.meta['filename']
于 2017-06-11T12:49:06.140 回答