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python - 如何从另一个函数中的一个函数访问数据

问题描述

我有一个网络抓取程序，在我的蜘蛛中，我需要访问我的 process_csv 函数中的信息，但我需要从我的 start_requests 函数中访问它。在我的 process_csv 函数中，我有一个变量“ip”，它从 CSV 文件中获取 ips，并将 http:// 添加到开头，将端口号添加到末尾，然后该函数继续将它们与需要的用户代理和 URL 配对被刮掉。无论如何，我需要在我的 start_requests 函数中访问这些 ip，因此我需要从我的 start_requests 函数中调用变量“ip”。我该怎么做？

SplashSpider.py

import csv
from scrapy.spiders import Spider
from scrapy_splash import SplashRequest
from ..items import GameItem

# process the csv file so the url + ip address + useragent pairs are the same as defined in the file # returns a list of dictionaries, example:
# [ {'url': 'http://www.starcitygames.com/catalog/category/Rivals%20of%20Ixalan',
#    'ip': 'http://204.152.114.244:8050',
#    'ua': "Mozilla/5.0 (BlackBerry; U; BlackBerry 9320; en-GB) AppleWebKit/534.11"},
#    ...
# ]
def process_csv(csv_file):
    data = []
    reader = csv.reader(csv_file)
    next(reader)
    for fields in reader:
        if fields[0] != "":
            url = fields[0]
        else:
            continue # skip the whole row if the url column is empty
        if fields[1] != "":
            ip = "http://" + fields[1] + ":8050" # adding http and port because this is the needed scheme
        if fields[2] != "":
            useragent = fields[2]
        data.append({"url": url, "ip": ip, "ua": useragent})
    return data


class MySpider(Spider):
    name = 'splash_spider'  # Name of Spider

    # notice that we don't need to define start_urls
    # just make sure to get all the urls you want to scrape inside start_requests function

    # getting all the url + ip address + useragent pairs then request them
    def start_requests(self):

        # get the file path of the csv file that contains the pairs from the settings.py
        with open(self.settings["PROXY_CSV_FILE"], mode="r") as csv_file:
            # requests is a list of dictionaries like this -> {url: str, ua: str, ip: str}
            requests = process_csv(csv_file)

        for req in requests:
            # no need to create custom middlewares
            # just pass useragent using the headers param, and pass proxy using the meta param
            yield SplashRequest(url=req["url"], callback=self.parse, args={"wait": 3},
                                splash_url = ip,
                                headers={"User-Agent": req["ua"]},
                                meta={
                                    "proxy": (req["ip"]),
                                  })

    # Scraping
#    def parse(self, response):
#        item = GameItem()
#        for game in response.css("tr"):
#            # Card Name
#            item["card_name"] = game.css("a.card_popup::text").extract_first()
#            yield item

标签： pythonscrapysplash-screenscrapy-splash

解决方案

在您的process_csv函数中，您将附加一个名为datawith的列表{"url": url, "ip": ip, "ua": useragent}。所以你可以从函数中调用这些记录中的每一个，这将返回一个字典，然后你可以调用 ip。例子：

request_ip = process_csv(csv_file)[0]["ip"]- 将返回列表中第一条记录中的 ip。您可以运行一个循环来获取所有 ip 并将它们存储在一个列表中。

希望这可以帮助。

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