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

你试过driver.implicitly_wait吗?它就像驱动程序的一个设置，所以你只在会话中调用它一次，它基本上告诉驱动程序等待给定的时间，直到每个命令都可以执行。

driver = webdriver.Chrome()
driver.implicitly_wait(10)

因此，如果您设置等待时间为10秒，它将尽快执行命令，等待10秒后才放弃。我在类似的滚动场景中使用过这个，所以我不明白为什么它在您的情况下不起作用。希望这对你有帮助。

为了能够修复这个答案，我必须添加新的文本。确保在implicitly_wait中使用小写“w”。

找到以下3种方法:

请求处理

检查页面readyState(不可靠):

def page_has_loaded(self):
    self.log.info("Checking if {} page is loaded.".format(self.driver.current_url))
    page_state = self.driver.execute_script('return document.readyState;')
    return page_state == 'complete'

wait_for helper函数很好，但不幸的是，click_through_to_new_page处于竞速条件，在浏览器开始处理单击之前，我们设法在旧页面中执行脚本，而page_has_loaded直接返回true。

id

比较新页面id和旧页面id:

def page_has_loaded_id(self):
    self.log.info("Checking if {} page is loaded.".format(self.driver.current_url))
    try:
        new_page = browser.find_element_by_tag_name('html')
        return new_page.id != old_page.id
    except NoSuchElementException:
        return False

比较id可能不如等待过时的引用异常有效。

staleness_of

使用staleness_of方法:

@contextlib.contextmanager
def wait_for_page_load(self, timeout=10):
    self.log.debug("Waiting for page to load at {}.".format(self.driver.current_url))
    old_page = self.find_element_by_tag_name('html')
    yield
    WebDriverWait(self, timeout).until(staleness_of(old_page))

要了解更多细节，请查看Harry的博客。

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.

这里我使用了一个相当简单的形式:

from selenium import webdriver
browser = webdriver.Firefox()
browser.get("url")
searchTxt=''
while not searchTxt:
    try:    
      searchTxt=browser.find_element_by_name('NAME OF ELEMENT')
      searchTxt.send_keys("USERNAME")
    except:continue

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)