Не удается сделать скриншот с селеном, если я использую execute_script функцию() - Селен

Question

Я использую Python Selenium с PhantomJS

Моя цель, чтобы выделить границы всех доступных на веб-странице форм, а затем взять скриншот. Вот что я делаю

dcap = dict(DesiredCapabilities.PHANTOMJS) 
    dcap["phantomjs.page.settings.userAgent"] = ("Mozilla/5.0 (Macintosh; Intel Mac OS X 10_10_1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/38.0.2125.122 Safari/537.36") 

    self.driver = webdriver.PhantomJS(desired_capabilities=dcap, service_args=['--ignore-ssl-errors=true', '--ssl-protocol=any', '--web-security=false']) 
    self.driver.set_window_size(1024, 768) 
    self.driver.get(response.url) 
    self.driver.execute_script("document.getElementsByTagName('form').style['border'] = 'solid'; document.getElementsByTagName('form').style['border-color'] = 'red'") 

    self.driver.save_screenshot('test.png') # save a screenshot to disk 

Я получаю

Traceback (most recent call last): 
    File "/usr/local/lib/python2.7/dist-packages/Scrapy-1.2.1-py2.7.egg/scrapy/utils/defer.py", line 102, in iter_errback 
    yield next(it) 
    File "/usr/local/lib/python2.7/dist-packages/Scrapy-1.2.1-py2.7.egg/scrapy/spidermiddlewares/offsite.py", line 29, in process_spider_output 
    for x in result: 
    File "/usr/local/lib/python2.7/dist-packages/Scrapy-1.2.1-py2.7.egg/scrapy/spidermiddlewares/referer.py", line 22, in <genexpr> 
    return (_set_referer(r) for r in result or()) 
    File "/usr/local/lib/python2.7/dist-packages/Scrapy-1.2.1-py2.7.egg/scrapy/spidermiddlewares/urllength.py", line 37, in <genexpr> 
    return (r for r in result or() if _filter(r)) 
    File "/usr/local/lib/python2.7/dist-packages/Scrapy-1.2.1-py2.7.egg/scrapy/spidermiddlewares/depth.py", line 58, in <genexpr> 
    return (r for r in result or() if _filter(r)) 
    File "/home/mani/legibot/scrapy_app/spiders/spider.py", line 63, in parse_page 
    self.driver.execute_script("document.getElementsByTagName('form').style['border'] = 'solid'; document.getElementsByTagName('form').style['border-color'] = 'red'") 
    File "/usr/local/lib/python2.7/dist-packages/selenium/webdriver/remote/webdriver.py", line 465, in execute_script 
    'args': converted_args})['value'] 
    File "/usr/local/lib/python2.7/dist-packages/selenium/webdriver/remote/webdriver.py", line 236, in execute 
    self.error_handler.check_response(response) 
    File "/usr/local/lib/python2.7/dist-packages/selenium/webdriver/remote/errorhandler.py", line 192, in check_response 
    raise exception_class(message, screen, stacktrace) 
WebDriverException: Message: {"errorMessage":"undefined is not an object (evaluating 'document.getElementsByTagName('form').style['border'] = 'solid'')","request":{"headers":{"Accept":"application/json","Accept-Encoding":"identity","Connection":"close","Content-Length":"211","Content-Type":"application/json;charset=UTF-8","Host":"127.0.0.1:55507","User-Agent":"Python-urllib/2.7"},"httpVersion":"1.1","method":"POST","post":"{\"sessionId\": \"b09db570-b01b-11e6-b238-a5d1487ff8e7\", \"args\": [], \"script\": \"document.getElementsByTagName('form').style['border'] = 'solid'; document.getElementsByTagName('form').style['border-color'] = 'red'\"}","url":"/execute","urlParsed":{"anchor":"","query":"","file":"execute","directory":"/","path":"/execute","relative":"/execute","port":"","host":"","password":"","user":"","userInfo":"","authority":"","protocol":"","source":"/execute","queryKey":{},"chunks":["execute"]},"urlOriginal":"/session/b09db570-b01b-11e6-b238-a5d1487ff8e7/execute"}} 
Screenshot: available via screen 

Если я закомментировать execute_script функцию, то скриншот получает успешно сохранен.

Я пытался добавить задержку между помощью execute_script и принимая скриншоте как

self.driver.execute_script("document.getElementsByTagName('form').style['border'] = 'solid'; document.getElementsByTagName('form').style['border-color'] = 'red'") 
time.sleep(2) 
self.driver.save_screenshot('test.png') # save a screenshot to disk 

Но ту же ошибку.

Answer 1

getElementsByTagName() возвращает несколько элементов - массив элементов (на самом деле HTMLCollection) - который не имеет style свойство, которое объясняет ошибку undefined is not an object.

Либо получить один элемент по индексу:

document.getElementsByTagName('form')[0].style['border'] = 'solid'; // ... 

Или найдите элемент с помощью селена, а затем передать его в качестве аргумента в сценарий:

form = self.driver.find_element_by_tag_name("form") 
self.driver.execute_script("arguments[0].style['border'] = 'solid'; arguments[0].style['border-color'] = 'red';", form) 

И это масштабируется до нескольких также:

forms = self.driver.find_elements_by_tag_name("form") 
for form in forms: 
    self.driver.execute_script(""" 
     var form = arguments[0]; 
     form.style['border'] = 'solid'; 
     form.style['border-color'] = 'red'; 
    """, form)