如何在scrapy中提交表单？

Question

我尝试使用 scrapy 完成登录并收集我的项目提交计数。这是代码。

from scrapy.item import Item, Field
from scrapy.http import FormRequest
from scrapy.spider import Spider
from scrapy.utils.response import open_in_browser


class GitSpider(Spider):
    name = "github"
    allowed_domains = ["github.com"]
    start_urls = ["https://www.github.com/login"]

    def parse(self, response):
        formdata = {'login': 'username',
                'password': 'password' }
        yield FormRequest.from_response(response,
                                        formdata=formdata,
                                        clickdata={'name': 'commit'},
                                        callback=self.parse1)

    def parse1(self, response):
        open_in_browser(response)

运行代码后

scrapy runspider github.py

它应该显示表单的结果页面，这应该是登录失败的同一页面，因为用户名和密码是假的。但是它向我显示了 search page. The log file is located in pastebin

代码应该如何修复？提前致谢。

Answer 1

使用 webdriver 的解决方案。

from selenium import webdriver
from selenium.webdriver.common.action_chains import ActionChains
import time
from scrapy.contrib.spiders import CrawlSpider

class GitSpider(CrawlSpider):

    name = "gitscrape"
    allowed_domains = ["github.com"]
    start_urls = ["https://www.github.com/login"]

    def __init__(self):
        self.driver = webdriver.Firefox()

    def parse(self, response):
        self.driver.get(response.url)
        login_form = self.driver.find_element_by_name('login')
        password_form = self.driver.find_element_by_name('password')
        commit = self.driver.find_element_by_name('commit')
        login_form.send_keys("yourlogin")
        password_form.send_keys("yourpassword")
        actions = ActionChains(self.driver)
        actions.click(commit)
        actions.perform()
        # by this point you are logged to github and have access 
        #to all data in the main menù
        time.sleep(3)
        self.driver.close()

Answer 2

您的问题是 FormRequest.from_response() 使用了不同的形式 - "search form"。但是，您希望它改用 "log in form"。提供一个 formnumber 参数：

yield FormRequest.from_response(response,
                                formnumber=1,
                                formdata=formdata,
                                clickdata={'name': 'commit'},
                                callback=self.parse1)

这是应用更改后我在浏览器中打开的内容（使用 "fake" 用户）：

Answer 3

使用 "formname" 参数也有效：

yield FormRequest.from_response(response,
                            formname='Login',
                            formdata=formdata,
                            clickdata={'name': 'commit'},
                            callback=self.parse1)

如何在scrapy中提交表单？

How to submit a form in scrapy?

python

forms

scrapy

web-scraping