Question

我正在尝试使用Firefox Webdriver ， Selenium 和 Python

问题在于，当Selenium试图获取主页时，它会崩溃，并显示以下错误消息：

/usr/bin/python2.7 /home/ghasemi/PycharmProjects/phorcys_watcher/main.py http://apps.evozi.com/apk-downloader/?id=com.instagram.android Traceback (most recent call last): File "/home/ghasemi/PycharmProjects/phorcys_watcher/main.py", line 7, in <module> content = google_play_download("com.instagram.android") File "/home/ghasemi/PycharmProjects/phorcys_watcher/collector.py", line 20, in google_play_download browser.get("https://apps.evozi.com/apk-downloader/?id=" + app_page_id) File "/usr/local/lib/python2.7/dist-packages/selenium/webdriver/remote/webdriver.py", line 268, in get self.execute(Command.GET, {'url': url}) File "/usr/local/lib/python2.7/dist-packages/selenium/webdriver/remote/webdriver.py", line 256, in execute self.error_handler.check_response(response) File "/usr/local/lib/python2.7/dist-packages/selenium/webdriver/remote/errorhandler.py", line 194, in check_response raise exception_class(message, screen, stacktrace) selenium.common.exceptions.WebDriverException: Message: Reached error page: about:neterror?e=unknownProtocolFound&u=httpss%3A//www.adnetworkperformance.com/script/java.php%3Foption%3Drotateur%26r%3D411313%26treqn%3D1025813717%26runauction%3D1%26crr%3D168ce9d76b1a6695b12e%2CwcwHrNzGnshFns2PnM3bbcwGW8xLz-mNycwuvZjurZja3MzJfMxG_9xMX4wYns7a2YxHvshBL9xe3shbjN2J7umN6umNm-mNuN2czNw723956800778f24b2db6%26rtid%3D595b6ecb8ac19%26cbrandom%3D0.7519066097934798%26cbtitle%3DAPK%2520Downloader%2520%255BLatest%255D%2520Download%2520Directly%2520%257C%2520Chrome%2520Extension%2520v3%2520%28Evozi%2520Official%29%26cbiframe%3D0%26cbWidth%3D1280%26cbHeight%3D717%26cbdescription%3DDownload%2520APKs%2520Directly%2520From%2520Google%2520Play%2520To%2520Your%2520Computer%2520With%2520APK%2520Downloader%2520Extension%2520For%2520Google%2520Chrome%26cbkeywords%3D%26cbref%3D&c=&f=regular&d=Firefox%20doesn%E2%80%99t%20know%20how%20to%20open%20this%20address%2C%20because%20one%20of%20the%20following%20protocols%20%28httpss%29%20isn%E2%80%99t%20associated%20with%20any%20program%20or%20is%20not%20allowed%20in%20this%20context.

此行引发异常： browser.get（ “https://apps.evozi.com/apk-downloader/?id=com.instagram.android”）

如上所示，此错误的来源是selenium尝试下载的页面中的错误链接。我找到导致此错误的框架：

<iframe width="468" height="60" marginwidth="0" marginheight="0" vspace="0" hspace="0" allowtransparency="true" allowfullscreen="true" style="border: medium none; padding: 0; margin: 0;" sandbox="allow-scripts allow-forms allow-popups allow-popups-to-escape-sandbox allow-pointer-lock allow-same-origin" id="595b6e88086f8" frameborder="0" src="httpss://www.adnetworkperformance.com/script/java.php?option=rotateur&r=411313&treqn=501505383&runauction=1&crr=fc25086f39dc3c58bbdbGJTJyVGZh9Gbud3bk1yawFmRyUSbvNmLpp3b2VmLzBHchZkMlYkMlE0MlMHc0RHa2dfe473f7c304fd8fb65&rtid=595b6e88086f8&cbrandom=0.6676681852413189&cbtitle=APK%20Downloader%20%5BLatest%5D%20Download%20Directly%20%7C%20Chrome%20Extension%20v3%20(Evozi%20Official)&cbiframe=0&cbWidth=1522&cbHeight=741&cbdescription=Download%20APKs%20Directly%20From%20Google%20Play%20To%20Your%20Computer%20With%20APK%20Downloader%20Extension%20For%20Google%20Chrome&cbkeywords=&cbref=" scrolling="no"></iframe>

正如您所看到的，网页开发人员错误地放置了httpss而不是https（两次！）。

我该如何处理这个问题？

更新

我的刮刀：

import requests from lxml import html from pyvirtualdisplay import Display from selenium import webdriver def google_play_download(app_page_id): browser = webdriver.Firefox() browser.get("https://apps.evozi.com/apk-downloader/?id=" + app_page_id) browser.find_element_by_css_selector(".btn.btn-primary.btn-lg.btn-block").click() apk_link = browser.find_element_by_css_selector(".btn.btn-success.btn-block").get_attribute('href') browser.quit() for rnd in range(5): resp = requests.get(apk_link) if resp.headers['Content-Length'] == str(len(resp.content)): return resp.content if __name__ == "__main__": content = google_play_download("com.instagram.android") f = open('./file', 'wb') f.write(content) f.close() [1]: https://apps.evozi.com/apk-downloader/

Answer 1

一个解决方案是url-parsing funktion，你在driver.get（url）之前调用每个url

def url_parser(url):
    if 'httpss' in url:
        url = url.replace('httpss','https')
    return url

然后你就这样使用它

url = url_parser(url)
driver.get(url)

Answer 2

你快到了......

import requests
import time
from selenium import webdriver

def google_play_download(app_page_id):
    browser = webdriver.Chrome()
    browser.get("https://apps.evozi.com/apk-downloader/?id=" + app_page_id)
    browser.find_element_by_css_selector(".btn.btn-primary.btn-lg.btn-block").click()
    time.sleep(10)

    apk_link = browser.find_element_by_css_selector(".btn.btn-success.btn-block").get_attribute('href')
    browser.quit()
    for rnd in range(5):
        resp = requests.get(apk_link)
        if resp.headers['Content-Length'] == str(len(resp.content)):
            return resp.content


if __name__ == "__main__":
    content = google_play_download("com.instagram.android")
    f = open('file.apk', 'wb')
    f.write(content)
    f.close()

Answer 3

The parser is not able to extract the proper url in this line apk_link = browser.find_element_by_css_selector(".btn.btn-success.btn-block").get_attribute('href')

As when you print the url its showing https://apps.evozi.com/apk-downloader/?id=com.instagram.android

Just change the line to

apk_link = browser.find_element_by_css_selector(".btn.btn-success.btn-block")
ele=apk_link.get_attribute('href')

for rnd in range(5):
      resp = requests.get(ele)
      if resp.headers['Content-Length'] == str(len(resp.content)):
            return resp.content

The code will work without error

Selenium返回unknownProtocolFound错误

3 个答案: