Selenium - 使用 element.click() 时出现过时元素引用异常

Question

错误 Output：

selenium.common.exceptions.StaleElementReferenceException: Message: The element reference of <span class="a-size-medium a-color-base a-text-normal"> is stale; either 
the element is no longer attached to the DOM, it is not in the current frame context, or the document has been refreshed

我试图用 selenium 做以下事情：

1. go 到亚马逊
2. 使用 python 自动搜索无聊的东西
3. 点击第一个产品标题
4. 如果价格标签元素，在产品网站上，打印它，然后 go 返回上一页
5. 如果不可用，go 返回上一页
6. go 到下一个产品标题并从步骤 4 开始重复

但是它在click()处失败

代码：

def experiment2():
    browser = webdriver.Firefox()
    browser.get("https://www.amazon.com/")
    searchelement = browser.find_element_by_css_selector("#twotabsearchtextbox")
    searchelement.send_keys('automate the boring stuff with python')
    searchelement.submit()
    time.sleep(5)
    elements = browser.find_elements_by_css_selector("span.a-color-base.a-text-normal")
    for element in elements:
        element.click()
        try:
            element = browser.find_element_by_css_selector("span.a-size-medium:nth-child(2)")
            print(element.text)
        except:
            browser.back()
            time.sleep(2)
            continue
        browser.back()
        time.sleep(2)

Answer 1

由于您使用的是 driver.back() 它刷新了页面并且您捕获的元素不再附加到页面。您需要再次重新分配元素。

试试下面的代码

def experiment2():
    browser = webdriver.Firefox()
    browser.get("https://www.amazon.com/")
    searchelement = browser.find_element_by_css_selector("#twotabsearchtextbox")
    searchelement.send_keys('automate the boring stuff with python')
    searchelement.submit()
    time.sleep(5)
    elements = browser.find_elements_by_css_selector("span.a-color-base.a-text-normal")
    for element in range(len(elements)):
        elements = browser.find_elements_by_css_selector("span.a-color-base.a-text-normal")
        elements[element].click()
        try:
            element = browser.find_element_by_css_selector("span.a-size-medium:nth-child(2)")
            print(element.text)
        except:
            browser.back()
            time.sleep(2)
            continue
        browser.back()
        time.sleep(2)