Python Selenium - 從LeafletJS彈出框和NoSuchElementException中獲取文本

Question

如何根據leafletjs從地圖彈出框中獲取文本。該url is this。我也得到這個錯誤：NoSuchElementException

如果我搜索了包裹並輸入了相關信息，如下所示，那麼我想要獲得包含class_name的所有文本：'leaflet-popup-content'？

# Creates an instance driver object... 
driver = webdriver.Chrome() 

# load the url above 
driver.get(url) 

# ============= 
# Find and fill SEARCH BOX by id.... 
driver.find_element_by_id('searchBox').send_keys('1083CX') 

# Send the form by clicking on the searcht botton... 
driver.find_element_by_id('searchButton').click() 

driver.find_element_by_id('listElementContent0').click() 
# driver.find_element_by_class_name('content').click() 

# ============= 
txt = driver.find_element_by_class_name('leaflet-popup-content').text() 
print (txt) 

這個問題在What is the best way to avoid NoSuchElementException in Selenium?的問題使用Java，我不明白。我正在使用Python，並且對所有這些都是新的？

Answer 1

這是一個時機的問題，你需要讓瀏覽器加載搜索結果，並在彈出的內容：

from selenium.webdriver.common.by import By 
from selenium.webdriver.support.ui import WebDriverWait 
from selenium.webdriver.support import expected_conditions as EC 

# ... 

wait = WebDriverWait(driver, 10) 

driver.find_element_by_id('searchBox').send_keys('1083CX') 
driver.find_element_by_id('searchButton').click() 

wait.until(EC.visibility_of_element_located((By.ID, 'listElementContent0'))).click() 

# wait for popup to appear and contain data 
wait.until(EC.visibility_of_element_located((By.CSS_SELECTOR, '.leaflet-popup-content b'))) 

popup = driver.find_element_by_class_name("leaflet-popup-content") 
print(popup.text) 

對我的作品和版畫：

Kadastrale aanduiding: ASD30AK02554G0000 
Kadastrale grootte (m2) : 930 
aanvullende gegevens bij het kadaster aanvragen