另外，您可以检查DOM是否没有更多的修改，而不是向下滚动100次(在页面底部是AJAX惰性加载的情况下)

def scrollDown(driver, value):
    driver.execute_script("window.scrollBy(0,"+str(value)+")")

# Scroll down the page
def scrollDownAllTheWay(driver):
    old_page = driver.page_source
    while True:
        logging.debug("Scrolling loop")
        for i in range(2):
            scrollDown(driver, 500)
            time.sleep(2)
        new_page = driver.page_source
        if new_page != old_page:
            old_page = new_page
        else:
            break
    return True

如果您试图滚动并找到页面上的所有项目。您可以考虑使用以下方法。这是其他人在这里提到的一些方法的组合。它帮我完成了任务:

while True:
    try:
        driver.execute_script("window.scrollTo(0, document.body.scrollHeight);")
        driver.implicitly_wait(30)
        time.sleep(4)
        elem1 = WebDriverWait(driver, 30).until(EC.presence_of_all_elements_located((By.CSS_SELECTOR, "element-name")))
        len_elem_1 = len(elem1)
        print(f"A list Length {len_elem_1}")
        driver.execute_script("window.scrollTo(0, document.body.scrollHeight);")
        driver.implicitly_wait(30)
        time.sleep(4)
        elem2 = WebDriverWait(driver, 30).until(EC.presence_of_all_elements_located((By.CSS_SELECTOR, "element-name")))
        len_elem_2 = len(elem2)
        print(f"B list Length {len_elem_2}")
        if len_elem_1 == len_elem_2:
            print(f"final length = {len_elem_1}")
            break
    except TimeoutException:
            print("Loading took too much time!")

从硒/ webdriver /支持/ wait.py

driver = ...
from selenium.webdriver.support.wait import WebDriverWait
element = WebDriverWait(driver, 10).until(
    lambda x: x.find_element_by_id("someId"))

正如David Cullen的回答中提到的，我总是看到这样的建议:

element_present = EC.presence_of_element_located((By.ID, 'element_id'))
WebDriverWait(driver, timeout).until(element_present)

对于我来说，很难找到所有可以与By一起使用的定位器，所以我认为在这里提供列表会很有用。 根据Ryan Mitchell的Web Scraping with Python:

ID Used in the example; finds elements by their HTML id attribute CLASS_NAME Used to find elements by their HTML class attribute. Why is this function CLASS_NAME not simply CLASS? Using the form object.CLASS would create problems for Selenium's Java library, where .class is a reserved method. In order to keep the Selenium syntax consistent between different languages, CLASS_NAME was used instead. CSS_SELECTOR Finds elements by their class, id, or tag name, using the #idName, .className, tagName convention. LINK_TEXT Finds HTML tags by the text they contain. For example, a link that says "Next" can be selected using (By.LINK_TEXT, "Next"). PARTIAL_LINK_TEXT Similar to LINK_TEXT, but matches on a partial string. NAME Finds HTML tags by their name attribute. This is handy for HTML forms. TAG_NAME Finds HTML tags by their tag name. XPATH Uses an XPath expression ... to select matching elements.

我挣扎了一点，让这个工作，因为它没有为我工作的预期。任何还在努力让它工作的人，可以检查一下。

我想等待一个元素出现在网页上，然后再继续我的操作。

我们可以使用WebDriverWait(driver, 10,1).until()，但catch是until()期望一个函数，它可以执行一段时间的超时提供(在我们的情况下是10)每1秒。所以保持它如下对我有用。

element_found = wait_for_element.until(lambda x: x.find_element_by_class_name("MY_ELEMENT_CLASS_NAME").is_displayed())

下面是until()在幕后所做的事情

def until(self, method, message=''):
        """Calls the method provided with the driver as an argument until the \
        return value is not False."""
        screen = None
        stacktrace = None

        end_time = time.time() + self._timeout
        while True:
            try:
                value = method(self._driver)
                if value:
                    return value
            except self._ignored_exceptions as exc:
                screen = getattr(exc, 'screen', None)
                stacktrace = getattr(exc, 'stacktrace', None)
            time.sleep(self._poll)
            if time.time() > end_time:
                break
        raise TimeoutException(message, screen, stacktrace)

nono = driver.current_url
driver.find_element(By.XPATH,"//button[@value='Send']").click()
  while driver.current_url == nono:
      pass
print("page loaded.")