我有一个文本文档,每行都有一个URL。我希望在一个新标签页中打开每个URL。这是我到目前为止的内容:
tabs = 0
f = open('links.txt', 'r', encoding='utf-8')
for line in f:
url = line
driver.execute_script("window.open(url, 'new_window')")
sleep(7) # time to let tge page load
tabs = tabs + 1 # to keep track of how many tabs I have
这给了我错误:
回溯(最近一次通话最后一次):文件“ scraper.py”,第100行,在 driver.execute_script(“ window.open(url,'new_window')”)文件“ C:\ Python37 \ lib \ site-packages \ selenium \ webdriver \ remote \ webdriver.py”, 第635行,在execute_script中 'args':convert_args})['value']文件“ C:\ Python37 \ lib \ site-packages \ selenium \ webdriver \ remote \ webdriver.py”, 第320行,执行 self.error_handler.check_response(响应)文件“ C:\ Python37 \ lib \ site-packages \ selenium \ webdriver \ remote \ errorhandler.py”, 第242行,在check_response中 引发exception_class(消息,屏幕,堆栈跟踪)selenium.common.exceptions.WebDriverException:消息:未知错误: 网址未定义
我已经尝试了所有方法,但无法正常工作。有任何想法吗?
答案 0 :(得分:0)
您试图在url
中错误地传递execute_script
变量。改为-
tabs = 0
f = open('links.txt', 'r', encoding='utf-8')
for line in f:
url = line
driver.execute_script("window.open(arguments[0])", url)
sleep(7) # time to let tge page load
tabs = tabs + 1 # to keep track of how many tabs I have
arguments[0]
是指您的网址。