使用Selenium（python）拍摄多个URL的屏幕截图 [英] Take screenshot of multiple URLs using selenium (python)

问题描述

我正在尝试使用python硒库对多个网站进行截图。

I am trying to take a screenshot of multiple websites using python selenium library.

在这里，我有一系列类似

Here I have an array of website like

data = array of website [ 'google.com', 'youtube.com'... ]

如果网站加载时间太长，我希望程序开始打开下一个网站。

If a website takes too long to load, I want the program starts open next websites.

但是代码不符合我的预期...

But this code doesn't do what I expected...

browser = webdriver.Chrome('/Users/wk/Desktop/checkSafeContent/chromedriver')

for index, url in enumerate(data):
    browser.set_page_load_timeout(30)    
    try:
        browser.get('http://'+data[index])
    except:
        print("takes too long")
        browser.quit()
        browser = webdriver.Chrome('/Users/wk/Desktop/checkSafeContent/chromedriver')
    else:
        browser.maximize_window()
        browser.implicitly_wait(20)
        # where images saved
        browser.save_screenshot('/.../'+str(index)+'.png')

browser.quit()

我认为我应该使用browser.close（），但我不知道该怎么做。

I think I should use browser.close(), but I don't know exactly how.

推荐答案

您应该花一些时间阅读所使用的不同语句的文档。您错误地使用了几种。

You should spend some time reading the docs for the different statements that you are using. You are using several incorrectly.

我认为这会起作用。一个问题可能是，如果页面加载时间过长，将不允许浏览器使用 browser.get（）导航到新页面。您可以尝试发送ESC密钥或通过谷歌搜索找到的许多其他选项之一。

I think this will work. One issue may be that if the page loads long, the browser will not be allowed to navigate to a new page with browser.get(). You might try sending an ESC key or one of the many other options you can find by googling.

我将网站添加到输入时间过长消息中，这样您就可以知道哪些未及时完成加载。

I added the site to the "took too long" message so you would know which ones didn't finish loading in time.

browser = webdriver.Chrome('/Users/wk/Desktop/checkSafeContent/chromedriver')
browser.set_page_load_timeout(30)    
browser.maximize_window()

for index, url in enumerate(data):
    try:
        browser.get('http://' + data[index])
    except:
        print(data[index] + ' took too long')
    else:
        # where images saved
        browser.save_screenshot('/.../' + str(index) + '.png')

browser.quit()

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