从Google MAPS中提取评分和评论数量

Question

我仍然是初学者，正在尝试提取图卢兹市 100 个最佳花园的评级（例如 4.1 颗星）和评论数量。我已经编写了下面的代码，但是我仍然无法提取所需的信息。

有人可以帮忙吗

提前致谢， 伊利亚斯。

from selenium import webdriver
from selenium.webdriver.chrome.options import Options
from selenium.webdriver.common.by import By
from selenium.webdriver.common.keys import Keys
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver import ActionChains
from selenium.webdriver.support.ui import WebDriverWait
import time

chrome_options = Options()
chrome_options.add_experimental_option("detach", True)

driver = webdriver.Chrome(options=chrome_options)
actionChains = ActionChains(driver)
wait = WebDriverWait(driver, 20)

def wait_for_element_location_to_be_stable(element):
    initial_location = element.location
    previous_location = initial_location
    start_time = time.time()
    while time.time() - start_time < 1:
        current_location = element.location
        if current_location != previous_location:
            previous_location = current_location
            start_time = time.time()
        time.sleep(0.4)

driver.get("https://www.google.com/")
driver.find_element(By.ID, "L2AGLb").click()

driver.get("https://www.google.com/maps")

# Wait for the page to load and display the search box
time.sleep(3)

# Input the search query for "jardins in toulouse" and press Enter
search_box = driver.find_element(By.ID, "searchboxinput")
search_box.send_keys("jardins in toulouse")
search_box.send_keys(Keys.RETURN)

# Wait for the search results to load
time.sleep(5)

# Updated XPath for the rating
rating_xpath = "/html/body/div[3]/div[8]/div[9]/div/div/div[1]/div[2]/div/div[1]/div/div/div[2]/div[1]"

# Get the elements based on the below XPath locator
results = driver.find_elements(By.XPATH, "//a[@class='hfpxzc']")

break_condition = False
focus_element = driver.find_element(By.ID, 'zero-input')
while not break_condition:
    temp = results[-1]
    actionChains.scroll_to_element(results[-1]).perform()
    actionChains.move_to_element(focus_element).click().perform()
    for i in range(3):
        actionChains.send_keys(Keys.ARROW_DOWN).perform()
        time.sleep(0.5)
    wait_for_element_location_to_be_stable(temp)

    results = wait.until(EC.presence_of_all_elements_located((By.XPATH, "//a[@class='hfpxzc']")))
    if results[-1] == temp:
        break_condition = True

# Loop through the results and print name and rating
for i, result in enumerate(results[:100], start=1):
    name = result.get_attribute('aria-label')
    rating_element = result.find_element(By.XPATH, rating_xpath)
    rating = rating_element.text if rating_element else "Rating not available"
    print(f"Result {i}: {name} - Rating: {rating}")

# Close the browser
driver.quit()

我知道这里有非常熟练的人员可以提供帮助......

Answer 1

我很高兴您使用我的解决方案来向下滚动 Google 地图。 :)

在当前情况下，您使用 DOM 根的绝对 xpath 获得评级，这是行不通的。该 xpath 仅从上下文根通向第一个评级元素。 并且

result

变量本身没有您定义为评级 xpath 的元素（它不是上下文根）。

评级放置在卡片容器中。卡片容器根是

a

元素的父级。所以，你定义

..

构建定位器的第二步是查找评级元素。它有独特的类别

MW4etd

因此，结果元素中的评级定位器是

..//*[@class='MW4etd']

要截断非评级结果，可以从

result

获取元素列表并检查它是否大于0。

for i, result in enumerate(results[:100], start=1):
    name = result.get_attribute('aria-label')
    ratings = result.find_elements(By.XPATH, "..//*[@class='MW4etd']")
    rating = ratings[0].text if len(ratings) > 0 else "Rating not available"
    print(f"Result {i}: {name} - Rating: {rating}")