我已经用Scrapy成功地抓取了一个网站上的图像数据,并将图像保存在一个文件夹中。但我想将图像的路径名保存在mysql数据库中。
就像下面的爬行器的结果一样,我想将数据“路径”转发到管道,但我不知道如何选择它
'images': [{'checksum': '75873dcc0944e29787525197648aa1a6',
'path': 'full/91e6d13e3ad32def287f98199c8bbe1915c71773.jpg',
'url': 'https://cdn.sindonews.net/dyn/620/content/2019/08/05/12/1426977/masa-kampanye-jadwal-pemilu-hingga-e-voting-jadi-isu-revisi-uu-pemilu-qoQ.jpg'}],我的pipelines.py:
import mysql.connector
class SkripsiPipeline(object):
def __init__(self):
self.create_connection()
# dispatcher.connect(self.close_spider, signals.close_spider)
# self.create_table()
def create_connection(self):
self.conn = mysql.connector.connect(
host = '127.0.0.1',
password = '',
user = 'root',
database = 'news'
)
self.curr = self.conn.cursor()
def process_item(self, item, spider):
self.store_db(item)
return item
def store_db(self,item):
self.curr.execute("INSERT INTO news_tb (url, title, author, time, crawl_time, image_urls, images, content) values (%s,%s,%s,%s,%s,%s,%s,%s)",(
item['url'][0],
item['title'][0],
item['author'][0],
item['time'][0],
item['crawl_time'][0],
item['image_urls'][0],
item['content'][0]
))
self.conn.commit()我的items.py
import scrapy
class SkripsiItem(scrapy.Item):
url = scrapy.Field()
title = scrapy.Field()
author = scrapy.Field()
time = scrapy.Field()
crawl_time = scrapy.Field()
image_urls = scrapy.Field()
images = scrapy.Field()
content = scrapy.Field()我想将图像路径的名称保存到数据库中。任何熟悉这个问题的人,请让我知道。谢谢你。
发布于 2019-08-05 19:58:18
如果图片列表是你的项目的一部分,你可以像这样选择它:item['images'][0]['path']。
您可以通过更改store_db方法将其添加到管道,如下所示:
def store_db(self,item):
path = item['images'][0]['path'].split('/')[1]
self.curr.execute("INSERT INTO news_tb (url, title, author, time, crawl_time, image_urls, images, content, path) values (%s,%s,%s,%s,%s,%s,%s,%s,%s)",(
item['url'][0],
item['title'][0],
item['author'][0],
item['time'][0],
item['crawl_time'][0],
item['image_urls'][0],
item['content'][0],
path
))
self.conn.commit()https://stackoverflow.com/questions/57357984
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