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I use selenium(python) to retrieve some data from WOS top papers, but when I use click() to open the sub link, I can only open the first url

My task is to open each url from the following website and retrieve some evaluation data for each essay. I have located the element successfully, which means I get 10 element. However, when selenium began to imitate human to click the url, it can only open the first link of ten links.
https://esi.clarivate.com/DocumentsAction.action
HTML:
The code is as followed.
import time
from selenium import webdriver
driver=webdriver.Chrome('/usr/local/bin/chromedriver')
driver.get('https://esi.clarivate.com/IndicatorsAction.action?Init=Yes&SrcApp=IC2LS&SID=H3-M1jrs4mSS2O3WTFbtdrUJugtDvogGRIM-18x2dx2B1ubex2Bo9Y5F6ZPQtUZbfUAx3Dx3Dp1StTsneXx2B7vu85UqXoaoQx3Dx3D-03Ff2gF3hTJGBPDScD1wSwx3Dx3D-cLUx2FoETAVeN3rTSMreq46gx3Dx3D')
#add filter-> research fields-> "clinical medicine"
target = driver.find_element_by_id("ext-gen1065")
time.sleep(1)
target.click()
time.sleep(1)
n = driver.window_handles
driver.switch_to.window(n[-1])
links=driver.find_elements_by_class_name("docTitle")
length=len(links)
for i in range(0,length):
item=links[i]
item.click()
time.sleep(1)
handles=driver.window_handles
index_handle=driver.current_window_handle
for handle in handles:
if handle != index_handle:
driver.switch_to.window(handle)
else:
continue
time.sleep(1)
u1=driver.find_elements_by_class_name("large-number")[2].text
u2=driver.find_elements_by_class_name("large-number")[3].text
print(u1,u2)
print("\n")
driver.close()
time.sleep(1)
driver.switch_to_window(index_handle)
driver.quit()
print("————finished————")
The error page:
And I try to find out the problem by testing these code:
links=driver.find_elements_by_class_name("docTitle")
length=len(links)
print(length)
print(links[1].text)
#links[0].click()
links[1].click()
The result is:
which means it had already find the element, but failed to open it.(when using links[0].text, it works fine.)
Any idea about this?

For Loops while using selenium for webscraping Python

I am attempting to web-scrape info off of the following website: https://www.axial.net/forum/companies/united-states-family-offices/
I am trying to scrape the description for each family office, so "https://www.axial.net/forum/companies/united-states-family-offices/"+insert_company_name" are the pages I need to scrape.
So I wrote the following code to test the program for just one page:
from bs4 import BeautifulSoup as soup
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
driver = webdriver.Chrome('insert_path_here/chromedriver')
driver.get("https://network.axial.net/company/ansaco-llp")
page_source = driver.page_source
soup2 = soup(page_source,"html.parser")
soup2.findAll('axl-teaser-description')[0].text
This works for the single page, as long as the description doesn't have a "show full description" drop down button. I will save that for another question.
I wrote the following loop:
#Note: Lst2 has all the names for the companies. I made sure they match the webpage
lst3=[]
for key in lst2[1:]:
driver.get("https://network.axial.net/company/"+key.lower())
page_source = driver.page_source
for handle in driver.window_handles:
driver.switch_to.window(handle)
word_soup = soup(page_source,"html.parser")
if word_soup.findAll('axl-teaser-description') == []:
lst3.append('null')
else:
c = word_soup.findAll('axl-teaser-description')[0].text
lst3.append(c)
print(lst3)
When I run the loop, all of the values come out as "null", even the ones without "click for full description" buttons.
I edited the loop to instead print out "word_soup", and the page is different then if I had run it without a loop and does not have the description text.
I don't understand why a loop would cause that but apparently it does. Does anyone know how to fix this problem?

Found solution. pause the program for 3 seconds after driver.get:
import time
lst3=[]
for key in lst2[1:]:
driver.get("https://network.axial.net/company/"+key.lower())
time.sleep(3)
page_source = driver.page_source
word_soup = soup(page_source,"html.parser")
if word_soup.findAll('axl-teaser-description') == []:
lst3.append('null')
else:
c = word_soup.findAll('axl-teaser-description')[0].text
lst3.append(c)
print(lst3)

I see that the page uses javascript to generate the text meaning it doesn't show up in the page source, which is weird but ok. I don't quite understand why you're only iterating through and switching to all the instances of Selenium you have open, but you definitely won't find the description in the page source / beautifulsoup.
Honestly, I'd personally look for a better website if you can, otherwise, you'll have to try it with selenium which is inefficient and horrible.

Facing issues while scraping data from a table using python with selenium

I've written a script using python in combination with selenium to parse table from a target page which can be reached out following some steps I've tried to describe below for the clarity. It does reach the destination but at the time of scraping data from that table It throws an error showing in the console "Unable to locate element". I tried with online xpath tester to see if it is wrong but I found that the xpath I've used in my script for "td_data" is right. I suppose, what I'm missing here is beyond my knowledge. Hope there is somebody to take a look into it and provide me with a workaround.
Btw, the site link is given in my script.
Link to see the html contents for the table: "https://www.dropbox.com/s/kaom5qzk78xndqn/Partial%20Html%20content%20for%20the%20table.txt?dl=0"
Steps to reach the target page which my script is able to maintain:
Selecting "I've read and understand above"
Putting this keyword "pump" in the inputbox located right below "Select medical devices".
Selecting the checkbox "Devices found for "pump".
Finally, pressing the search button
Script I've tried with so far:
from selenium import webdriver
import time
driver = webdriver.Chrome()
driver.get('http://apps.tga.gov.au/Prod/devices/daen-entry.aspx')
driver.find_element_by_id('disclaimer-accept').click()
time.sleep(5)
driver.find_element_by_id('medicine-name').send_keys('pump')
time.sleep(8)
driver.find_element_by_id('medicines-header-text').click()
driver.find_element_by_id('submit-button').click()
time.sleep(7)
for item in driver.find_elements_by_xpath('//div[#class="table-responsive"]'):
for tr_data in item.find_elements_by_xpath('.//tr'):
td_data = tr_data.find_element_by_xpath('.//span[#class="hovertext"]//a')
print(td_data.text)
driver.close()

Why don't you just do this:
from selenium import webdriver
import time
driver = webdriver.Chrome()
driver.get('http://apps.tga.gov.au/Prod/devices/daen-entry.aspx')
driver.find_element_by_id('disclaimer-accept').click()
time.sleep(5)
driver.find_element_by_id('medicine-name').send_keys('pump')
time.sleep(8)
driver.find_element_by_id('medicines-header-text').click()
driver.find_element_by_id('submit-button').click()
time.sleep(7)
for item in driver.find_elements_by_xpath(
'//table[#id]/tbody/tr/td[#class]/span[#class]/a[#id]'
):
print(item.text)
driver.close()
Output:
27233
27283
27288
27289
27390
27413
27441
27520
25445
27816
27866
27970
28033
28238
26999
28264
28407
28448
28437
28509
28524
28553
28647
28677
28646
Maybe you want to think about saving the page with driver.page_source, pull out the table, save it as a html file. Then use pandas from html to open the table into a dataframe

Python and Selenium: I am automating web scraping among pages. How can I loop by Next button?

I already written several lines of codes to pull url from this website.
http://www.worldhospitaldirectory.com/United%20States/hospitals
code is below:
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
import time
import csv
driver = webdriver.Firefox()
driver.get('http://www.worldhospitaldirectory.com/United%20States/hospitals')
url = []
pagenbr = 1
while pagenbr <= 115:
current = driver.current_url
driver.get(current)
lks = driver.find_elements_by_xpath('//*[#href]')
for ii in lks:
link = ii.get_attribute('href')
if '/info' in link:
url.append(link)
print('page ' + str(pagenbr) + ' is done.')
if pagenbr <=114:
elm = driver.find_element_by_link_text('Next')
driver.implicitly_wait(10)
elm.click()
time.sleep(2)
pagenbr += 1
ls = list(set(url))
with open('US_GeneralHospital.csv', 'wb') as myfile:
wr = csv.writer(myfile,quoting=csv.QUOTE_ALL)
for u in ls:
wr.writerow([u])
And it worked very well to pull each individual links from this website.
But the problem is I need to change the page number I need to loop by myself every time.
I want to let this code upgrade to iterate by calculating how many time it need. Not by manually inputting.
Thank you very much.

This is bad idea to hardcode the number of pages in your script. Try just to click "Next" button while it is enabled:
from selenium.common.exceptions import NoSuchElementException
while True:
try:
# do whatever you need to do on page
driver.find_element_by_xpath('//li[not(#class="disabled")]/span[text()="Next"]').click()
except NoSuchElementException:
break
This should allow you to execute page scraping until the last page reached
Also note that using lines current = driver.current_url and driver.get(current) makes no sense at all, so you might skip them

Scrape with BeautifulSoup from site that uses AJAX pagination using Python

I'm fairly new to coding and Python so I apologize if this is a silly question. I'd like a script that goes through all 19,000 search results pages and scrapes each page for all of the urls. I've got all of the scrapping working but can't figure out how to deal with the fact that the page uses AJAX to paginate. Usually I'd just make a loop with the url to capture each search result but that's not possible. Here's the page: http://www.heritage.org/research/all-research.aspx?nomobile&categories=report
This is the script I have so far:
with io.open('heritageURLs.txt', 'a', encoding='utf8') as logfile:
page = urllib2.urlopen("http://www.heritage.org/research/all-research.aspx?nomobile&categories=report")
soup = BeautifulSoup(page)
snippet = soup.find_all('a', attrs={'item-title'})
for a in snippet:
logfile.write ("http://www.heritage.org" + a.get('href') + "\n")
print "Done collecting urls"
Obviously, it scrapes the first page of results and nothing more.
And I have looked at a few related questions but none seem to use Python or at least not in a way that I can understand. Thank you in advance for your help.

For the sake of completeness, while you may try accessing the POST request and to find a way round to access to next page, like I suggested in my comment, if an alternative is possible, using Selenium will be quite easy to achieve what you want.
Here is a simple solution using Selenium for your question:
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
from time import sleep
# uncomment if using Firefox web browser
driver = webdriver.Firefox()
# uncomment if using Phantomjs
#driver = webdriver.PhantomJS()
url = 'http://www.heritage.org/research/all-research.aspx?nomobile&categories=report'
driver.get(url)
# set initial page count
pages = 1
with open('heritageURLs.txt', 'w') as f:
while True:
try:
# sleep here to allow time for page load
sleep(5)
# grab the Next button if it exists
btn_next = driver.find_element_by_class_name('next')
# find all item-title a href and write to file
links = driver.find_elements_by_class_name('item-title')
print "Page: {} -- {} urls to write...".format(pages, len(links))
for link in links:
f.write(link.get_attribute('href')+'\n')
# Exit if no more Next button is found, ie. last page
if btn_next is None:
print "crawling completed."
exit(-1)
# otherwise click the Next button and repeat crawling the urls
pages += 1
btn_next.send_keys(Keys.RETURN)
# you should specify the exception here
except:
print "Error found, crawling stopped"
exit(-1)
Hope this helps.