CODEWITHSUNDEEP
python-3.xselenium-webdriver
Tester P
Tester P

Reputation: 177

How to take screenshot of element using python3 and selenium

I want to take screenshot of the element. I tried with below mentioned code:

ElementIs=driver.find_element_by_xpath(".//body/img")
ElementIs.screenshot('abc.jpg')

but got error:

File "C:\Python34\lib\site-packages\selenium-2.47.1-y3.4.egg\selenium\webdriver\remote\webelement.py", line 448, in _execute   return self._parent.execute(command, params)
File "C:\Python34\lib\site-packages\selenium-2.47.1-y3.4.egg\selenium\webdriver\remote\webdriver.py", line 196, in execute self.error_handler.check_response(response)
File "C:\Python34\lib\site-packages\selenium-2.47.1-py3.4.egg\selenium\webdriver\remote\errorhandler.py", line 102, in check_responsevalue = json.loads(value_json)
File "C:\Python34\lib\json\__init__.py", line 318, in loads   return _default_decoder.decode(s)
File "C:\Python34\lib\json\decoder.py", line 343, in decode   obj, end = self.raw_decode(s, idx=_w(s, 0).end())
File "C:\Python34\lib\json\decoder.py", line 361, in raw_decode raise ValueError(errmsg("Expecting value", s, err.value)) from None ValueError: Expecting value: line 1 column 1 (char 0)

Upvotes: 0

Views: 4216

Answers (2)

Wictor Chaves
Wictor Chaves

Reputation: 1129

With Firefox you can:

driver = webdriver.Firefox()
qr = driver.find_element_by_css_selector('.css_class')
qr.screenshot("abc.jpg")

You can use class, id or tags in find_element_by_css_selector.

Upvotes: 1

Andrew Regan
Andrew Regan

Reputation: 5113

A couple of problems:

  1. find_elements_by_xpath returns a list, and you can't request a screenshot on a list. Did you mean find_element_by_xpath?
  2. In any case, with the exception of Microsoft Edge, no WebDriver actually implements element-based screenshots. You have to generate a full-window screenshot and crop it.

See my earlier answer on this topic for more background and workarounds.

Upvotes: 3

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