硒，存在于许多元素中的一种定位？

Question

在这种情况下使用any（）的正确方式是什么？或者更好的是，需要做什么才能使这种方法发挥作用？

假设Selenium.get('url')在调用此方法之前已经执行。

def wait_with_xpath_expectation(self, search_elements, timeout=6, max_attempts=3):
    """
    Selenium wait for an element(s) designated by Xpath to become available in the DOM. Useful for javascript AJAXy
    loading pages where content may be be rendered dynamically on the client side after page load appears complete.
    search_elements may be one Xpath as a string or many as a list. This allows for any of multiple elements on a
    page or pages to be determined to have loaded based on expectations.
    :param search_elements: string or list (strings converted to lists), Xpath(s)
    :param timeout: int, seconds
    :param max_attempts: int, time to wait before giving up on page loading
    :return: Boolean, True if page has loaded, False if all attempts have failed
    """

    # Page is not loaded yet
    loaded = False

    # Initialize attempt count
    attempt = 0

    # If only one element has been passed, ensure it is encapsulated by a list
    if type(search_elements) is str:
        search_elements = [search_elements]

    # Begin the polling loop
    while attempt < max_attempts:

        try:

            while loaded is False:
                # Create a list of expected elements using list comprehension
                expectations = [EC.presence_of_element_located((By.XPATH, element)) for element in search_elements]

                # Define wait
                wait = WebDriverWait(self.browser, timeout)

                # Execute
                wait.until(any(expectations))

                # Acknowledge load
                loaded = True

                print("Success: Page loaded based on expectations")

                # Exit returning True for load
                return loaded

        except TimeoutException as e:

            # Increment attempts
            attempt += 1

            # Check again if max attempts has not been exceeded
            while attempt < max_attempts:

                # Increase timeout by 20%
                timeout *= .2

                # Try again 
                continue

            # Print an error if max attempts is exceeded
            print("Error: Max load with expectations attempts exceeded,", e)

            # Exit returning False for load
            return loaded

Answer 1

您可以有一个预期条件类来等待预期条件的组合。这里有一个例子。

class wait_for_all(object):
    def __init__(self, methods):
        self.methods = methods

    def __call__(self, driver):
        try:
            for method in self.methods:
                if not method(driver):
                    return False
            return True
        except StaleElementReferenceException:
            return False

然后通过构建预期条件的数组并在相同的等待中检查所有条件来使用此方法。（为清晰起见，示例行被拆分。）

methods = []
methods.append(EC.visibility_of_element_located(BY.ID, "idElem1"))
methods.append(EC.visibility_of_element_located(BY.ID, "idElem2"))
method = wait_for_all(methods)
WebDriverWait(driver, 5).until(method)

这将执行一个五秒的等待，同时检查两个不同元素的可见性。