小编Abh*_*ath的帖子

我有一个URL(例如http://www.example.com/OtterBox-77-24444-Commuter-Series-Optimus/dp/B00A21KPEI/ref=pd_sim_cps_4),想要截取它并在我的网页上预览它.意思是,用户点击预览按钮,PhantomJS需要以PNG/JPEG格式预览网页

我也可以使用任何其他开源.

我有一个CSV文件,其中包含以下行:

"AAA", "BBB", "Test, Test", "CCC"
"111", "222, 333", "XXX", "YYY, ZZZ" 

等等 ...

我不想用双引号解析逗号.即.我的预期结果应该是

AAA
BBB
Test, Test
CCC

我的代码:

import csv
with open('values.csv', 'rb') as f:
    reader = csv.reader(f)
    for row in reader:
        print row

我尝试在python下使用csv包但没有运气.解析会爆炸所有逗号.

如果我遗失了什么,请告诉我

我无法抓取整个网站,Scrapy只是在表面爬行,我想爬得更深.谷歌搜索最后5-6小时,没有任何帮助.我的代码如下:

from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from scrapy.selector import HtmlXPathSelector
from scrapy.item import Item
from scrapy.spider import BaseSpider
from scrapy import log

class ExampleSpider(CrawlSpider):
    name = "example.com"
    allowed_domains = ["example.com"]
    start_urls = ["http://www.example.com/"]
    rules = [Rule(SgmlLinkExtractor(allow=()), 
                  follow=True),
             Rule(SgmlLinkExtractor(allow=()), callback='parse_item')
    ]
    def parse_item(self,response):
        self.log('A response from %s just arrived!' % response.url)

请帮忙 !!!!

谢谢,阿比拉姆