Python Selenium - 从 LeafletJS 弹出框和 NoSuchElementException 获取文本

Question

如何从基于leafletjs 的地图弹出框中获取文本。 url is this。我也收到此错误：NoSuchElementException

如果我搜索包裹并输入如下所示的相关信息，那么我想获取带有 class_name: 'leaflet-popup-content' 的框上的所有文本？

# 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)

What is the best way to avoid NoSuchElementException in Selenium? 的这个问题的分析器使用了 Java，我不明白。我正在使用 Python 并且对这一切都不熟悉？

Answer 1

这是时间问题，需要让浏览器加载搜索结果和弹窗内容：

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)

为我工作并打印：

Kadastrale aanduiding: ASD30AK02554G0000
Kadastrale grootte (m2) : 930
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