Navigate through content generated by Javascript using Python in Selernium? - javascript

I've written a script to test a process involving data input & several pages, but after writing it I've found the forms & main content to be generated from javascript.
The following is a snippet of the script I wrote, and after that initial link the content is generated by JS (its my first python script so excuse any mistakes);
from selenium import webdriver
from selenium.common.exceptions import NoSuchElementException
from selenium.webdriver.common.keys import Keys
from selenium.webdriver.common.action_chains import ActionChains
import time
browser = webdriver.Firefox()
browser.get('http://127.0.0.1:46727/?ajax=1')
assert "Home" in browser.title
# Find and click the Employer Database link
empDatabaseLink = browser.find_element_by_link_text('Employer Database')
click = ActionChains(browser).click(on_element = empDatabaseLink)
click.perform()
# Content loaded by the above link is generated by the JS
# Find and click the Add Employer button
addEmployerButton = browser.find_element_by_id('Add Employer')
addEmployer = ActionChains(browser).click(on_element = addEmployerButton)
addEmployer.perform()
browser.save_screenshot(r'images\Add_Employer_Form.png')
# Input Employer name
employerName = browser.find_element_by_id('name')
employerName.send_keys("Selenium")
browser.save_screenshot(r'images\Entered_Employer_Name.png')
# Move to next
nextButton = broswer.find_element_by_name('button_next')
moveForward = ActionChains(browser).click(on_element = nextButton)
# Move through various steps
# Then
# Move to Finish
moveForward = ActionChains(browser).click(on_element = nextButton)
How do you access page elements that aren't in the source? I've been looking around & found GetEval but not found anything that I can use :/

Well, to the people of the future, our above conversation appears to have lead to the conclusion that xpath is what mark was looking for. So remember to try xpath, and to use the Selenium IDE and Firebug to locate particularly obstinate page elements.

Related

Python Selenium Take a Screenshot Of An Whole Page Without Using Headless Mode

I need to take a screenshot of an element which is very long and not fit on the screen, I can use headless mode to do this but site doesn't allow me to do even with user-agent and other stuff.
But I can access the site with undetectedChromeDriver, so there's a extension to do this stuff called 'HTML Elements Screenshot'.
That extension will allow you to select element and take the screenshot of an whole element for you.
I automated that process with pyautogui and cv2 but I want to do it without this libraries. Is there any Javascript code to do it ? I did my research but can't find any useful. Thanks in advance
The codes I tried:
def save_screenshot(driver, path: str = 'screenshot.png') -> None:
input('Let it go when u ready.')
driver.switch_to.window(driver.window_handles[-1])
original_size = driver.get_window_size()
required_width = driver.execute_script('return document.body.parentNode.scrollWidth')
required_height = driver.execute_script('return document.body.parentNode.scrollHeight')
driver.set_window_size(required_width, required_height)
driver.save_screenshot(path) # has scrollbar
#driver.find_element_by_tag_name('body').screenshot(path) # avoids scrollbar
driver.set_window_size(original_size['width'], original_size['height'])
def saveScreenshot(driver,path: str="screenshot.png"):
input('Let it go when u ready.')
driver.switch_to.window(driver.window_handles[-1])
el = driver.find_element(By.TAG_NAME,"body")
el.screenshot(path)
driver.quit()
For python you can use pyppeteer.
For javascript you can use puppeteer
You can find the documentation here

Selenium for crawling javascript in python : using page_source [duplicate]

I want to scrape all the data of a page implemented by a infinite scroll. The following python code works.
for i in range(100):
driver.execute_script("window.scrollTo(0, document.body.scrollHeight);")
time.sleep(5)
This means every time I scroll down to the bottom, I need to wait 5 seconds, which is generally enough for the page to finish loading the newly generated contents. But, this may not be time efficient. The page may finish loading the new contents within 5 seconds. How can I detect whether the page finished loading the new contents every time I scroll down? If I can detect this, I can scroll down again to see more contents once I know the page finished loading. This is more time efficient.
The webdriver will wait for a page to load by default via .get() method.
As you may be looking for some specific element as #user227215 said, you should use WebDriverWait to wait for an element located in your page:
from selenium import webdriver
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.common.by import By
from selenium.common.exceptions import TimeoutException
browser = webdriver.Firefox()
browser.get("url")
delay = 3 # seconds
try:
myElem = WebDriverWait(browser, delay).until(EC.presence_of_element_located((By.ID, 'IdOfMyElement')))
print "Page is ready!"
except TimeoutException:
print "Loading took too much time!"
I have used it for checking alerts. You can use any other type methods to find the locator.
EDIT 1:
I should mention that the webdriver will wait for a page to load by default. It does not wait for loading inside frames or for ajax requests. It means when you use .get('url'), your browser will wait until the page is completely loaded and then go to the next command in the code. But when you are posting an ajax request, webdriver does not wait and it's your responsibility to wait an appropriate amount of time for the page or a part of page to load; so there is a module named expected_conditions.
Trying to pass find_element_by_id to the constructor for presence_of_element_located (as shown in the accepted answer) caused NoSuchElementException to be raised. I had to use the syntax in fragles' comment:
from selenium import webdriver
from selenium.common.exceptions import TimeoutException
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.common.by import By
driver = webdriver.Firefox()
driver.get('url')
timeout = 5
try:
element_present = EC.presence_of_element_located((By.ID, 'element_id'))
WebDriverWait(driver, timeout).until(element_present)
except TimeoutException:
print "Timed out waiting for page to load"
This matches the example in the documentation. Here is a link to the documentation for By.
Find below 3 methods:
readyState
Checking page readyState (not reliable):
def page_has_loaded(self):
self.log.info("Checking if {} page is loaded.".format(self.driver.current_url))
page_state = self.driver.execute_script('return document.readyState;')
return page_state == 'complete'
The wait_for helper function is good, but unfortunately click_through_to_new_page is open to the race condition where we manage to execute the script in the old page, before the browser has started processing the click, and page_has_loaded just returns true straight away.
id
Comparing new page ids with the old one:
def page_has_loaded_id(self):
self.log.info("Checking if {} page is loaded.".format(self.driver.current_url))
try:
new_page = browser.find_element_by_tag_name('html')
return new_page.id != old_page.id
except NoSuchElementException:
return False
It's possible that comparing ids is not as effective as waiting for stale reference exceptions.
staleness_of
Using staleness_of method:
#contextlib.contextmanager
def wait_for_page_load(self, timeout=10):
self.log.debug("Waiting for page to load at {}.".format(self.driver.current_url))
old_page = self.find_element_by_tag_name('html')
yield
WebDriverWait(self, timeout).until(staleness_of(old_page))
For more details, check Harry's blog.
As mentioned in the answer from David Cullen, I've always seen recommendations to use a line like the following one:
element_present = EC.presence_of_element_located((By.ID, 'element_id'))
WebDriverWait(driver, timeout).until(element_present)
It was difficult for me to find somewhere all the possible locators that can be used with the By, so I thought it would be useful to provide the list here.
According to Web Scraping with Python by Ryan Mitchell:
ID
Used in the example; finds elements by their HTML id attribute
CLASS_NAME
Used to find elements by their HTML class attribute. Why is this
function CLASS_NAME not simply CLASS? Using the form object.CLASS
would create problems for Selenium's Java library, where .class is a
reserved method. In order to keep the Selenium syntax consistent
between different languages, CLASS_NAME was used instead.
CSS_SELECTOR
Finds elements by their class, id, or tag name, using the #idName,
.className, tagName convention.
LINK_TEXT
Finds HTML tags by the text they contain. For example, a link that
says "Next" can be selected using (By.LINK_TEXT, "Next").
PARTIAL_LINK_TEXT
Similar to LINK_TEXT, but matches on a partial string.
NAME
Finds HTML tags by their name attribute. This is handy for HTML forms.
TAG_NAME
Finds HTML tags by their tag name.
XPATH
Uses an XPath expression ... to select matching elements.
From selenium/webdriver/support/wait.py
driver = ...
from selenium.webdriver.support.wait import WebDriverWait
element = WebDriverWait(driver, 10).until(
lambda x: x.find_element_by_id("someId"))
On a side note, instead of scrolling down 100 times, you can check if there are no more modifications to the DOM (we are in the case of the bottom of the page being AJAX lazy-loaded)
def scrollDown(driver, value):
driver.execute_script("window.scrollBy(0,"+str(value)+")")
# Scroll down the page
def scrollDownAllTheWay(driver):
old_page = driver.page_source
while True:
logging.debug("Scrolling loop")
for i in range(2):
scrollDown(driver, 500)
time.sleep(2)
new_page = driver.page_source
if new_page != old_page:
old_page = new_page
else:
break
return True
Have you tried driver.implicitly_wait. It is like a setting for the driver, so you only call it once in the session and it basically tells the driver to wait the given amount of time until each command can be executed.
driver = webdriver.Chrome()
driver.implicitly_wait(10)
So if you set a wait time of 10 seconds it will execute the command as soon as possible, waiting 10 seconds before it gives up. I've used this in similar scroll-down scenarios so I don't see why it wouldn't work in your case. Hope this is helpful.
To be able to fix this answer, I have to add new text. Be sure to use a lower case 'w' in implicitly_wait.
Here I did it using a rather simple form:
from selenium import webdriver
browser = webdriver.Firefox()
browser.get("url")
searchTxt=''
while not searchTxt:
try:
searchTxt=browser.find_element_by_name('NAME OF ELEMENT')
searchTxt.send_keys("USERNAME")
except:continue
Solution for ajax pages that continuously load data. The previews methods stated do not work. What we can do instead is grab the page dom and hash it and compare old and new hash values together over a delta time.
import time
from selenium import webdriver
def page_has_loaded(driver, sleep_time = 2):
'''
Waits for page to completely load by comparing current page hash values.
'''
def get_page_hash(driver):
'''
Returns html dom hash
'''
# can find element by either 'html' tag or by the html 'root' id
dom = driver.find_element_by_tag_name('html').get_attribute('innerHTML')
# dom = driver.find_element_by_id('root').get_attribute('innerHTML')
dom_hash = hash(dom.encode('utf-8'))
return dom_hash
page_hash = 'empty'
page_hash_new = ''
# comparing old and new page DOM hash together to verify the page is fully loaded
while page_hash != page_hash_new:
page_hash = get_page_hash(driver)
time.sleep(sleep_time)
page_hash_new = get_page_hash(driver)
print('<page_has_loaded> - page not loaded')
print('<page_has_loaded> - page loaded: {}'.format(driver.current_url))
How about putting WebDriverWait in While loop and catching the exceptions.
from selenium import webdriver
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
from selenium.common.exceptions import TimeoutException
browser = webdriver.Firefox()
browser.get("url")
delay = 3 # seconds
while True:
try:
WebDriverWait(browser, delay).until(EC.presence_of_element_located(browser.find_element_by_id('IdOfMyElement')))
print "Page is ready!"
break # it will break from the loop once the specific element will be present.
except TimeoutException:
print "Loading took too much time!-Try again"
You can do that very simple by this function:
def page_is_loading(driver):
while True:
x = driver.execute_script("return document.readyState")
if x == "complete":
return True
else:
yield False
and when you want do something after page loading complete,you can use:
Driver = webdriver.Firefox(options=Options, executable_path='geckodriver.exe')
Driver.get("https://www.google.com/")
while not page_is_loading(Driver):
continue
Driver.execute_script("alert('page is loaded')")
use this in code :
from selenium import webdriver
driver = webdriver.Firefox() # or Chrome()
driver.implicitly_wait(10) # seconds
driver.get("http://www.......")
or you can use this code if you are looking for a specific tag :
from selenium import webdriver
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
driver = webdriver.Firefox() #or Chrome()
driver.get("http://www.......")
try:
element = WebDriverWait(driver, 10).until(
EC.presence_of_element_located((By.ID, "tag_id"))
)
finally:
driver.quit()
Very good answers here. Quick example of wait for XPATH.
# wait for sizes to load - 2s timeout
try:
WebDriverWait(driver, 2).until(expected_conditions.presence_of_element_located(
(By.XPATH, "//div[#id='stockSizes']//a")))
except TimeoutException:
pass
I struggled a bit to get this working as that didn't worked for me as expected. anyone who is still struggling to get this working, may check this.
I want to wait for an element to be present on the webpage before proceeding with my manipulations.
we can use WebDriverWait(driver, 10, 1).until(), but the catch is until() expects a function which it can execute for a period of timeout provided(in our case its 10) for every 1 sec. so keeping it like below worked for me.
element_found = wait_for_element.until(lambda x: x.find_element_by_class_name("MY_ELEMENT_CLASS_NAME").is_displayed())
here is what until() do behind the scene
def until(self, method, message=''):
"""Calls the method provided with the driver as an argument until the \
return value is not False."""
screen = None
stacktrace = None
end_time = time.time() + self._timeout
while True:
try:
value = method(self._driver)
if value:
return value
except self._ignored_exceptions as exc:
screen = getattr(exc, 'screen', None)
stacktrace = getattr(exc, 'stacktrace', None)
time.sleep(self._poll)
if time.time() > end_time:
break
raise TimeoutException(message, screen, stacktrace)
If you are trying to scroll and find all items on a page. You can consider using the following. This is a combination of a few methods mentioned by others here. And it did the job for me:
while True:
try:
driver.execute_script("window.scrollTo(0, document.body.scrollHeight);")
driver.implicitly_wait(30)
time.sleep(4)
elem1 = WebDriverWait(driver, 30).until(EC.presence_of_all_elements_located((By.CSS_SELECTOR, "element-name")))
len_elem_1 = len(elem1)
print(f"A list Length {len_elem_1}")
driver.execute_script("window.scrollTo(0, document.body.scrollHeight);")
driver.implicitly_wait(30)
time.sleep(4)
elem2 = WebDriverWait(driver, 30).until(EC.presence_of_all_elements_located((By.CSS_SELECTOR, "element-name")))
len_elem_2 = len(elem2)
print(f"B list Length {len_elem_2}")
if len_elem_1 == len_elem_2:
print(f"final length = {len_elem_1}")
break
except TimeoutException:
print("Loading took too much time!")
selenium can't detect when the page is fully loaded or not, but javascript can. I suggest you try this.
from selenium.webdriver.support.ui import WebDriverWait
WebDriverWait(driver, 100).until(lambda driver: driver.execute_script('return document.readyState') == 'complete')
this will execute javascript code instead of using python, because javascript can detect when page is fully loaded, it will show 'complete'. This code means in 100 seconds, keep tryingn document.readyState until complete shows.
nono = driver.current_url
driver.find_element(By.XPATH,"//button[#value='Send']").click()
while driver.current_url == nono:
pass
print("page loaded.")

Webscraping website that has a button to click

I am trying to webscrape a website that has multiple javascript rendered pages (https://openlibrary.ecampusontario.ca/catalogue/). I am able to get the content from the first page, but I am not sure how to get my script to click on the buttons on the subsequent pages to get that content. Here is my script.
import time
from bs4 import BeautifulSoup as soup
import requests
from selenium import webdriver
from selenium.webdriver.chrome.options import Options
import json
# The path to where you have your chrome webdriver stored:
webdriver_path = '/Users/rawlins/Downloads/chromedriver'
# Add arguments telling Selenium to not actually open a window
chrome_options = Options()
chrome_options.add_argument('--headless')
chrome_options.add_argument('--window-size=1920x1080')
# Fire up the headless browser
browser = webdriver.Chrome(executable_path = webdriver_path,
chrome_options = chrome_options)
# Load webpage
url = "https://openlibrary.ecampusontario.ca/catalogue/"
browser.get(url)
# to ensure that the page has loaded completely.
time.sleep(3)
data = []
# Parse HTML, close browser
page_soup = soup(browser.page_source, 'lxml')
containers = page_soup.findAll("div", {"class":"result-item tooltip"})
for container in containers:
item = {}
item['type'] = "Textbook"
item['title'] = container.find('h4', {'class' : 'textbook-title'}).text.strip()
item['author'] = container.find('p', {'class' : 'textbook-authors'}).text.strip()
item['link'] = "https://openlibrary.ecampusontario.ca/catalogue/" + container.find('h4', {'class' : 'textbook-title'}).a["href"]
item['source'] = "eCampus Ontario"
item['base_url'] = "https://openlibrary.ecampusontario.ca/catalogue/"
data.append(item) # add the item to the list
with open("js-webscrape-2.json", "w") as writeJSON:
json.dump(data, writeJSON, ensure_ascii=False)
browser.quit()
You do not have to actually click on any button. For example, to search for items with the keyword 'electricity', you navigate to the url
https://openlibrary-repo.ecampusontario.ca/rest/filtered-items?query_field%5B%5D=*&query_op%5B%5D=matches&query_val%5B%5D=(%3Fi)electricity&filters=is_not_withdrawn&offset=0&limit=10000
This will return a json string of items with the first item being:
{"items":[{"uuid":"6af61402-b0ec-40b1-ace2-1aa674c2de9f","name":"Introduction to Electricity, Magnetism, and Circuits","handle":"123456789/579","type":"item","expand":["metadata","parentCollection","parentCollectionList","parentCommunityList","bitstreams","all"],"lastModified":"2019-05-09 15:51:06.91","parentCollection":null,"parentCollectionList":null,"parentCommunityList":null,"bitstreams":null,"withdrawn":"false","archived":"true","link":"/rest/items/6af61402-b0ec-40b1-ace2-1aa674c2de9f","metadata":null}, ...
Now, to get that item, you use its uuid, and navigate to:
https://openlibrary.ecampusontario.ca/catalogue/item/?id=6af61402-b0ec-40b1-ace2-1aa674c2de9f
You can proceed like this for any interaction with that website (this is not always working for all websites, but it is working for your website).
To find out what are the urls that are navigated to when you click such and such button or enter text (what I did for the above urls), you can use fiddler.
I made a little script that can help you (selenium).
what this script does is "while the last page of the catalogue is not selected (in this case, contain 'selected' in it's class), i'll scrape , then click next"
while "selected" not in driver.find_elements_by_css_selector("[id='results-pagecounter-pages'] a")[-1].get_attribute("class"):
#your scraping here
driver.find_element_by_css_selector("[id='next-btn']").click()
There's probably a problem that you'll run into using this method, it doesn't wait for the results to load, but you can figure out what to do from here onwards.
Hope it helps

Parsing html from a javascript rendered url with python object

I would like to extract the market information from the following url and all of its subsequent pages:
https://uk.reuters.com/investing/markets/index/.FTSE?sortBy=&sortDir=&pn=1
I have successfully parsed the data that I want from the first page using some code from the following url:
https://impythonist.wordpress.com/2015/01/06/ultimate-guide-for-scraping-javascript-rendered-web-pages
I have also been able to parse out the url for the next page to feed into a loop in order to grab data from the next page. The problem is it crashes before the next page loads for a reason I don't fully understand.
I have a hunch that the class that I have borrowed from 'impythonist' may be causing the problem. I don't know enough object orientated programming to work out the problem. Here is my code, much of which is borrowed from the the url above:
import sys
from PyQt4.QtGui import *
from PyQt4.QtCore import *
from PyQt4.QtWebKit import *
from lxml import html
import re
from bs4 import BeautifulSoup
class Render(QWebPage):
def __init__(self, url):
self.app = QApplication(sys.argv)
QWebPage.__init__(self)
self.loadFinished.connect(self._loadFinished)
self.mainFrame().load(QUrl(url))
self.app.exec_()
def _loadFinished(self, result):
self.frame = self.mainFrame()
self.app.quit()
base_url='https://uk.reuters.com'
complete_next_page='https://uk.reuters.com/investing/markets/index/.FTSE?sortBy=&sortDir=&pn=1'
#LOOP TO RENDER PAGES AND GRAB DATA
while complete_next_page != '':
print ('NEXT PAGE: ',complete_next_page, '\n')
r = Render(complete_next_page) # USE THE CLASS TO RENDER JAVASCRIPT FROM PAGE
result = r.frame.toHtml() # ERROR IS THROWN HERE ON 2nd PAGE
# PARSE THE HTML
soup = BeautifulSoup(result, 'lxml')
row_data=soup.find('div', attrs={'class':'column1 gridPanel grid8'})
print (len(row_data))
# PARSE ALL ROW DATA
stripe_rows=row_data.findAll('tr', attrs={'class':'stripe'})
non_stripe_rows=row_data.findAll('tr', attrs={'class':''})
print (len(stripe_rows))
print (len(non_stripe_rows))
# PARSE SPECIFIC ROW DATA FROM INDEX COMPONENTS
#non_stripe_rows: from 4 to 18 (inclusive) contain data
#stripe_rows: from 2 to 16 (inclusive) contain data
i=2
while i < len(stripe_rows):
print('CURRENT LINE IS: ',str(i))
print(stripe_rows[i])
print('###############################################')
print(non_stripe_rows[i+2])
print('\n')
i+=1
#GETS LINK TO NEXT PAGE
next_page=str(soup.find('div', attrs={'class':'pageNavigation'}).find('li', attrs={'class':'next'}).find('a')['href']) #GETS LINK TO NEXT PAGE WORKS
complete_next_page=base_url+next_page
I have annotated the bits of code that I have written and understand but I don't really know what's going on in the 'Render' class enough to diagnose the error? Unless its something else?
Here is the error:
result = r.frame.toHtml()
AttributeError: 'Render' object has no attribute 'frame'
I don't need to keep the information in the class once I have parsed it out so I was thinking perhaps it could be cleared or reset somehow and then updated to hold the new url information from page 2:n but I have no idea how to do this?
Alternatively if anyone knows another way to grab this specific data from this page and the following ones then that would be equally helpful?
Many thanks in advance.
How about using selenium and phantomjs instead of PyQt.
You can easily get selenium by executing "pip install selenium".
If you use Mac you can get phantomjs by executing "brew install phantomjs".
If your PC is Windows use choco instead of brew, or Ubuntu use apt-get.
from selenium import webdriver
from bs4 import BeautifulSoup
base_url = "https://uk.reuters.com"
first_page = "/business/markets/index/.FTSE?sortBy=&sortDir=&pn=1"
browser = webdriver.PhantomJS()
# PARSE THE HTML
browser.get(base_url + first_page)
soup = BeautifulSoup(browser.page_source, "lxml")
row_data = soup.find('div', attrs={'class':'column1 gridPanel grid8'})
# PARSE ALL ROW DATA
stripe_rows = row_data.findAll('tr', attrs={'class':'stripe'})
non_stripe_rows = row_data.findAll('tr', attrs={'class':''})
print(len(stripe_rows), len(non_stripe_rows))
# GO TO THE NEXT PAGE
next_button = soup.find("li", attrs={"class":"next"})
while next_button:
next_page = next_button.find("a")["href"]
browser.get(base_url + next_page)
soup = BeautifulSoup(browser.page_source, "lxml")
row_data = soup.find('div', attrs={'class':'column1 gridPanel grid8'})
stripe_rows = row_data.findAll('tr', attrs={'class':'stripe'})
non_stripe_rows = row_data.findAll('tr', attrs={'class':''})
print(len(stripe_rows), len(non_stripe_rows))
next_button = soup.find("li", attrs={"class":"next"})
# DONT FORGET THIS!!
browser.quit()
I know the code above is not efficient (too slow I feel), but I think that it will bring you the results you desire. In addition, if the web page you want to scrape does not use Javascript, even PhantomJS and selenium are unnecessary. You can use the requests module. However, since I wanted to show you the contrast with PyQt, I used PhantomJS and Selenium in this answer.

How to get html with javascript rendered sourcecode by using selenium

I run a query in one web page, then I get result url. If I right click see html source, I can see the html code generated by JS. If I simply use urllib, python cannot get the JS code. So I see some solution using selenium. Here's my code:
from selenium import webdriver
url = 'http://www.archives.com/member/Default.aspx?_act=VitalSearchResult&lastName=Smith&state=UT&country=US&deathYear=2004&deathYearSpan=10&location=UT&activityID=9b79d578-b2a7-4665-9021-b104999cf031&RecordType=2'
driver = webdriver.PhantomJS(executable_path='C:\python27\scripts\phantomjs.exe')
driver.get(url)
print driver.page_source
>>> <html><head></head><body></body></html> Obviously It's not right!!
Here's the source code I need in right click windows, (I want the INFORMATION part)
</script></div><div class="searchColRight"><div id="topActions" class="clearfix
noPrint"><div id="breadcrumbs" class="left"><a title="Results Summary"
href="Default.aspx? _act=VitalSearchR ...... <<INFORMATION I NEED>> ...
to view the entire record.</p></div><script xmlns:msxsl="urn:schemas-microsoft-com:xslt">
jQuery(document).ready(function() {
jQuery(".ancestry-information-tooltip").actooltip({
href: "#AncestryInformationTooltip", orientation: "bottomleft"});
});
So my question is: How to get the information generated by JS?
You will need to get get the document via javascript you can use seleniums execute_script function
from time import sleep # this should go at the top of the file
sleep(5)
html = driver.execute_script("return document.getElementsByTagName('html')[0].innerHTML")
print html
That will get everything inside of the <html> tag
It's not necessary to use that workaround, you can use instead:
driver = webdriver.PhantomJS()
driver.get('http://www.google.com/')
html = driver.find_element_by_tag_name('html').get_attribute('innerHTML')
I have same problem about getting Javascript sourcecode from Internet, and I solved it using above Victory's suggestion.
*First: execute_script
driver=webdriver.Chrome()
driver.get(urls)
innerHTML = driver.execute_script("return document.body.innerHTML")
#print(driver.page_source)
*Second: parse html using beautifulsoup (You can Downloaded beautifulsoup by pip command)
import bs4 #import beautifulsoup
import re
from time import sleep
sleep(1) #wait one second
root=bs4.BeautifulSoup(innerHTML,"lxml") #parse HTML using beautifulsoup
viewcount=root.find_all("span",attrs={'class':'short-view-count style-scope yt-view-count-renderer'}) #find the value which you need.
*Third: print out the value you need
for span in viewcount:
print(span.string)
*Full code
from selenium import webdriver
import lxml
urls="http://www.archives.com/member/Default.aspx?_act=VitalSearchResult&lastName=Smith&state=UT&country=US&deathYear=2004&deathYearSpan=10&location=UT&activityID=9b79d578-b2a7-4665-9021-b104999cf031&RecordType=2"
driver = webdriver.PhantomJS()
##driver=webdriver.Chrome()
driver.get(urls)
innerHTML = driver.execute_script("return document.body.innerHTML")
##print(driver.page_source)
import bs4
import re
from time import sleep
sleep(1)
root=bs4.BeautifulSoup(innerHTML,"lxml")
viewcount=root.find_all("span",attrs={'class':'short-view-count style-scope yt-view-count-renderer'})
for span in viewcount:
print(span.string)
driver.quit()
I am thinking that you are getting the source code before the JavaScript has rendered the dynamic HTML.
Initially try putting a few seconds sleep between the navigate and get page source.
If this works, then you can change to a different wait strategy.
You try Dryscrape this browser is fully supported heavy js codes try it i hope it work for you
I met the same problem and finally solved by desired_capabilities.
from selenium import webdriver
from selenium.webdriver.common.proxy import Proxy
from selenium.webdriver.common.proxy import ProxyType
proxy = Proxy(
{
'proxyType': ProxyType.MANUAL,
'httpProxy': 'ip_or_host:port'
}
)
desired_capabilities = webdriver.DesiredCapabilities.PHANTOMJS.copy()
proxy.add_to_capabilities(desired_capabilities)
driver = webdriver.PhantomJS(desired_capabilities=desired_capabilities)
driver.get('test_url')
print driver.page_source

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