我想刮取无限滚动实现的页面的所有数据。下面的python代码可以工作。
for i in range(100):
driver.execute_script("window.scrollTo(0, document.body.scrollHeight);")
time.sleep(5)
这意味着每当我向下滚动到底部时，我都需要等待5秒，这通常足以让页面完成加载新生成的内容。但是，这可能并不省时。页面可能在5秒内完成新内容的加载。如何在每次向下滚动时检测页面是否完成了新内容的加载?如果我能检测到这一点，一旦我知道页面完成加载，我就可以再次向下滚动以查看更多内容。这样更节省时间。
当前回答
把WebDriverWait放在While循环中并捕获异常如何?
from selenium import webdriver
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
from selenium.common.exceptions import TimeoutException
browser = webdriver.Firefox()
browser.get("url")
delay = 3 # seconds
while True:
try:
WebDriverWait(browser, delay).until(EC.presence_of_element_located(browser.find_element_by_id('IdOfMyElement')))
print "Page is ready!"
break # it will break from the loop once the specific element will be present.
except TimeoutException:
print "Loading took too much time!-Try again"
其他回答
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.")
在代码中使用:
from selenium import webdriver
driver = webdriver.Firefox() # or Chrome()
driver.implicitly_wait(10) # seconds
driver.get("http://www.......")
或者你可以使用这段代码，如果你正在寻找一个特定的标签:
from selenium import webdriver
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
driver = webdriver.Firefox() #or Chrome()
driver.get("http://www.......")
try:
element = WebDriverWait(driver, 10).until(
EC.presence_of_element_located((By.ID, "tag_id"))
)
finally:
driver.quit()
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!")
回答得很好。等待XPATH的快速示例。
# wait for sizes to load - 2s timeout
try:
WebDriverWait(driver, 2).until(expected_conditions.presence_of_element_located(
(By.XPATH, "//div[@id='stockSizes']//a")))
except TimeoutException:
pass
