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[2019年9月4日]原网站反爬措施更新了, 代码更改如下

原网站反爬虫机制做了更改. 代码也需要相应更改(到2019年9月4日为止,更改两个地方即可绕过反爬机制)
①将AoiSolaSpider.py中的allowed_domains = ["www.mm131.com"]变为allowed_domains = ["www.mm131.com", "www.mm131.net"] (这里解决的是content被过滤的问题)
②将middlewares.py下AoisolasSpiderMiddleware类中process_request函数的内容整个换成: request.headers['referer'] = "http://www.mm131.com/?zzaqkey=4087969942"
(绕过防盗链)

下载失败

一直提示WARNING: File (code: 403): Error downloading file from *** referred in ***
WARNING: Dropped: Item contains no images,好像没有一张下载成功过,win10+python3.6+scrapy1.5

ModuleNotFoundError: No module named 'AoiSolas'错误

代码完全复制,但是总会出现
Traceback (most recent call last):
File "D:/zart/Aoisolas/Aoisolas/spiders/AoiSolaSpider.py", line 13, in
from AoiSolas.items import AoisolasItem
ModuleNotFoundError: No module named 'AoiSolas'
这样的错误,爬虫跑不起来
python3.7版本
前面几个项目都试了一下都可以,就这个出现这个错误,水平有限,找不出原因,求大佬指教

No module named 'PIL'

display error:
from scrapy.pipelines.images import ImagesPipeline
File "D:\Anaconda3\envs\my_env\lib\site-packages\scrapy\pipelines\images.py", line 15, in
from PIL import Image
ModuleNotFoundError: No module named 'PIL'

pip install pillow

子类重写def item_completed(self, results, item, info),可以实现文件重命名功能

from scrapy.pipelines.images import ImagesPipeline
from scrapy import Request
from ImageSpider.settings import IMAGES_STORE as images_store
import os

class ImagespiderPipeline(ImagesPipeline):

def get_media_requests(self, item, info):
    # 循环每一张图片地址下载,若传过来的不是集合则无需循环直接yield
    for image_url in item['imgurl']:
        yield Request(image_url)

# def file_path(self, request, response=None, info=None):
#     # 重命名,若不重写这函数,图片名为哈希,就是一串乱七八糟的名字
#     image_guid = request.url.split('/')[-1]  # 提取url前面名称作为图片名。
#     return image_guid

# def item_completed(self, results, item, info):
# 	#重命名文件,并把默认路径D:\ImageSpider\full\*图片 
# 	#修改为D:\ImageSpider\*.jpg,提取item['imgurl']中url前面名称作为图片名
# 	#功能上类似file_path
# 	image_path = [x["path"] for ok, x in results if ok]
# 	for i in range(len(image_path)):
# 		os.rename(images_store+'/'+image_path[i],images_store+'/'+item['imgurl'][i].split('/')[-1])

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