2017-02-14 461 views
1

我试图从python中的seroium在chromedriver中的标记中获取文本。我的代码正确运行,但是当没有应该得到的标记时,脚本会发出错误并停止。错误:Python - 为引发异常处理exception_class(消息,屏幕,堆栈跟踪)

raise exception_class(message, screen, stacktrace) 
selenium.common.exceptions.NoSuchElementException: Message: no such element: Unable to locate element: {"method":"xpath","selector":"//span[@class="section-facts-description-text"]"} 
    (Session info: chrome=56.0.2924.87) 
    (Driver info: chromedriver=2.27.440174(e97a722caafc2d3a8b807ee115bfb307f7d2cfd9),platform=Windows NT 10.0.14393 x86_64) 

我想使异常,所以当硒找不到标签,它会打印一条消息,而不是停止脚本。我的代码:

import os 
import sys 
from selenium import webdriver 

def findLocation(location): 
    browser = webdriver.Chrome() 
    print "please wait, I'll show you where "+location+" is" 
    browser.get("https://www.google.co.id/maps/place/"+location+"/") 
    try: 
     elem = browser.find_element_by_xpath('//span[@class="section-facts-description-text"]') 
     desc = elem.text 
     print str(desc) 
    except:  
     print "There's no description about this location" 

我不知道我应该用什么异常来处理这个错误。如果我只使用except它会打印“没有这个位置描述”太甚至有跨度标签

回答

0

只需添加到您的导入部分:

from selenium.common.exceptions import NoSuchElementException 

而且尝试/除部分如下:

try: 
    # do stuff 

except NoSuchElementException as e: 
    print e 
0

以外使用

except Exception: print "There's no description about this location"

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