Develop Reference

selenium stale element reference: element is not attached to the page document error

I have an e-commerce page and there are multiple products on a page. I need to click the link of a product then return on the main page and click the link of the next product, but when I return, the elements can't be found anymore.
Path = "C:\Program Files (x86)\chromedriver.exe"
driver = webdriver.Chrome(Path)
driver.get("https://www.emag.ro/")
search_bar = driver.find_element_by_id("searchboxTrigger")
search_bar.send_keys("laptopuri")
search_bar.send_keys(Keys.RETURN)
main = None
try:
main = WebDriverWait(driver, 10).until(
EC.presence_of_element_located((By.ID, "main-container"))
)
print("Page loaded,main retrived succesfully")
print(" ")
except:
driver.quit()
products = main.find_elements_by_css_selector("div.card-item.js-product-data")
for product in products:
raw_name = product.text
raw_price = product.find_element_by_css_selector("p.product-new-price").text
link = product.find_element_by_tag_name("a")
#clicking the link
link.click()
spec_page = None
try:
spec_page = WebDriverWait(driver, 10).until(
EC.presence_of_element_located((By.CLASS_NAME, "col-md-12"))
)
except:
driver.quit()
print(spec_page)
driver.back()
After the first iteration, I get the following error :
selenium.common.exceptions.StaleElementReferenceException: Message: stale element reference: element is not attached to the page document on line raw_name = product.text,basically at the beginning of the loop.
I assume the page is not loading properly or something like that, I tried using time.sleep before going through the loop but nothing

When you are writing driver.back(), it will go back to previous page and by the time it will reach to original page all the defined elements will become stale. You need to redefined them like below :-
This should handle the exception.
products = len(main.find_elements_by_css_selector("div.card-item.js-product-data"))
j = 0
for product in range(products):
elements = main.find_elements_by_css_selector("div.card-item.js-product-data")
raw_name = elements[j].text
raw_price = elements[j].find_element_by_css_selector("p.product-new-price").text
link = elements[j].find_element_by_tag_name("a")
# clicking the link
link.click()
spec_page = None
try:
spec_page = WebDriverWait(driver, 10).until(
EC.presence_of_element_located((By.CLASS_NAME, "col-md-12"))
)
except:
driver.quit()
print(spec_page)
j = j + 1
driver.back()

Scrolling a particular web element using Selenium python

I'm trying to scrape company's jobs offer from linkedin. I need to scroll a section in the page (with an inner scrollbar). I have been trying this :
1.
scroll_active = WebDriverWait(driver, 40).until(EC.presence_of_element_located((By.CSS_SELECTOR, "body > div.application-outlet > div.authentication-outlet > div.job-search-ext > div > div > section.jobs-search__left-rail > div > div > ul")))
scroll_active.location_once_scrolled_into_view
while driver.find_element_by_tag_name('div'):
driver.execute_script("window.scrollTo(0, document.body.scrollHeight);")
Divs=driver.find_element_by_tag_name('div').text
if 'End of Results' in Divs:
print 'end'
break
else:
continue
Need to extract 'href'

If any one facing that, I wish this could help, you just have to choose well the element that you want to scroll
my_xpath = WebDriverWait(driver, 40).until(EC.presence_of_element_located((By.XPATH, "/html/body/div[8]/div[3]/div[3]/div/div/section[1]/div/div")))
driver.execute_script('arguments[0].scrollTop = arguments[0].scrollHeight', my_xpath)

Why do need to scroll here?
seems like you can get all of the element by command:
elements = driver.find_elements(By.XPATH, "//a[#class='result-card__full-card-link']")
and looks like:
import time
from selenium import webdriver
from selenium.webdriver.common.by import By
driver = webdriver.Chrome()
driver.get('https://www.linkedin.com/jobs/search/?f_C=1110%2C12800%2C5115950%2C3165553%2C603115%2C10916%2C8331%2C3297950%2C8238%2C5509188%2C3093%2C2625246%2C1112%2C947572%2C11018069%2C407323&geoId=92000000')
time.sleep(3)
def element_present():
try:
driver.find_element(By.XPATH, "//button[#class='infinite-scroller__show-more-button infinite-scroller__show-more-button--visible']")
except Exception:
return False
return True
while not element_present():
driver.execute_script("window.scrollTo(0, document.body.scrollHeight);")
elements = driver.find_elements(By.XPATH, "//a[#class='result-card__full-card-link']")
hrefs = [el.get_attribute('href') for el in elements]
print(hrefs)
print(len(hrefs))
driver.quit()
might I missed smth, but seems like it works well as well

In selenium how to find out the exact number of XPATH links with different ids?

With Python3 and selenium I want to automate the search on a public information site. In this site it is necessary to enter the name of a person, then select the spelling chosen for that name (without or with accents or name variations), access a page with the list of lawsuits found and in this list you can access the page of each case.
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
from selenium.common.exceptions import TimeoutException, NoSuchElementException
from selenium.webdriver.common.keys import Keys
import time
import re
Name that will be searched
name = 'JOSE ROBERTO ARRUDA'
Create path, search start link, and empty list to store information
firefoxPath="/home/abraji/Documentos/Code/geckodriver"
link = 'https://ww2.stj.jus.br/processo/pesquisa/?aplicacao=processos.ea'
processos = []
Call driver and go to first search page
driver = webdriver.Firefox(executable_path=firefoxPath)
driver.get(link)
Position cursor, fill and click
WebDriverWait(driver, 5).until(EC.element_to_be_clickable((By.CSS_SELECTOR, '#idParteNome'))).click()
time.sleep(1)
driver.find_element_by_xpath('//*[#id="idParteNome"]').send_keys(name)
time.sleep(6)
WebDriverWait(driver, 5).until(EC.element_to_be_clickable((By.CSS_SELECTOR, '#idBotaoPesquisarFormularioExtendido'))).click()
Mark all spelling possibilities for searching
WebDriverWait(driver, 5).until(EC.element_to_be_clickable((By.CSS_SELECTOR, '#idBotaoMarcarTodos'))).click()
WebDriverWait(driver, 5).until(EC.element_to_be_clickable((By.CSS_SELECTOR, '#idBotaoPesquisarMarcados'))).click()
time.sleep(1)
Check how many pages of data there are - to be used in "for range"
capta = driver.find_element_by_xpath('//*[#id="idDivBlocoPaginacaoTopo"]/div/span/span[2]').text
print(capta)
paginas = int(re.search(r'\d+', capta).group(0))
paginas = int(paginas) + 1
print(paginas)
Capture routine
for acumula in range(1, paginas):
# Fill the field with the page number and press enter
driver.find_element_by_xpath('//*[#id="idDivBlocoPaginacaoTopo"]/div/span/span[2]/input').send_keys(acumula)
driver.find_element_by_xpath('//*[#id="idDivBlocoPaginacaoTopo"]/div/span/span[2]/input').send_keys(Keys.RETURN)
time.sleep(2)
# Captures the number of processes found on the current page - qt
qt = driver.find_element_by_xpath('//*[#id="idDivBlocoMensagem"]/div/b').text
qt = int(qt) + 2
print(qt)
# Iterate from found number of processes
for item in range(2, qt):
# Find the XPATH of each process link - start at number 2
vez = '//*[#id="idBlocoInternoLinhasProcesso"]/div[' + str(item) + ']/span[1]/span[1]/span[1]/span[2]/a'
print(vez)
# Access the direct link and click
element = WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.XPATH, vez)))
element.click()
# Run tests to get data
try:
num_unico = driver.find_element_by_xpath('//*[#id="idProcessoDetalhesBloco1"]/div[6]/span[2]/a').text
except NoSuchElementException:
num_unico = "sem_numero_unico"
try:
nome_proc = driver.find_element_by_xpath('//*[#id="idSpanClasseDescricao"]').text
except NoSuchElementException:
nome_proc = "sem_nome_encontrado"
try:
data_autu = driver.find_element_by_xpath('//*[#id="idProcessoDetalhesBloco1"]/div[5]/span[2]').text
except NoSuchElementException:
data_autu = "sem_data_encontrada"
# Fills dictionary and list
dicionario = {"num_unico": num_unico,
"nome_proc": nome_proc,
"data_autu": data_autu
}
processos.append(dicionario)
# Return a page to click on next process
driver.execute_script("window.history.go(-1)")
# Close driver
driver.quit()
In this case I captured the number of link pages (3) and the total number of links (84). So my initial idea was to do the "for" three times and within them split the 84 links
The direct address of each link is in XPATH (//*[#id="idBlocoInternoLinhasProcesso"]/div[41]/span[1]/span[1]/span[1]/span[2]/a) which I replace with the "item" to click
For example, when it arrives at number 42 I have an error because the first page only goes up to 41
My problem is how to go to the second page and then restart only "for" secondary
I think the ideal would be to know the exact number of links on each of the three pages
Anyone have any ideas?

Code below is "Capture routine":
wait = WebDriverWait(driver, 20)
#...
while True:
links = wait.until(EC.presence_of_all_elements_located((By.XPATH, "//span[contains(#class,'classSpanNumeroRegistro')]")))
print("links len", len(links))
for i in range(1, len(links) + 1):
# Access the direct link and click
.until(EC.element_to_be_clickable((By.XPATH, f"(//span[contains(#class,'classSpanNumeroRegistro')])[{i}]//a"))).click()
# Run tests to get data
try:
num_unico = driver.find_element_by_xpath('//*[#id="idProcessoDetalhesBloco1"]/div[6]/span[2]/a').text
except NoSuchElementException:
num_unico = "sem_numero_unico"
try:
nome_proc = driver.find_element_by_xpath('//*[#id="idSpanClasseDescricao"]').text
except NoSuchElementException:
nome_proc = "sem_nome_encontrado"
try:
data_autu = driver.find_element_by_xpath('//*[#id="idProcessoDetalhesBloco1"]/div[5]/span[2]').text
except NoSuchElementException:
data_autu = "sem_data_encontrada"
# Fills dictionary and list
dicionario = {"num_unico": num_unico,
"nome_proc": nome_proc,
"data_autu": data_autu
}
processos.append(dicionario)
# Return a page to click on next process
driver.execute_script("window.history.go(-1)")
# wait.until(EC.presence_of_element_located((By.CLASS_NAME, "classSpanPaginacaoImagensDireita")))
next_page = driver.find_elements_by_css_selector(".classSpanPaginacaoProximaPagina")
if len(next_page) == 0:
break
next_page[0].click()

You can try run the loop until next button is present on the screen. the logic will look like this,
try:
next_page = driver.find_element_by_class_name('classSpanPaginacaoProximaPagina')
if(next_page.is_displayed()):
next_page.click()
except NoSuchElementException:
print('next page does not exists')

python selenium: how to visit a web page many times on the same page

I'm using selenium with python to test my web server. Here is my test code:
i = 0
msg = 'abc'
while i < 10:
driver = webdriver.Chrome()
driver.get("http://www.example.com")
txt = driver.find_element_by_id('input-text')
txt.clear()
txt.send_keys(msg)
btn = driver.find_element_by_id('input-search')
btn.click()
driver.quit()
i += 1
The code works well except only one thing: it executes Chrome, do the test and close it for each time of loop. Obviously it's not necessary. What I need is simply to execute Chrome only one time and do many requests. I've tried as below but it doesn't work:
i = 0
msg = 'abc'
driver = webdriver.Chrome()
while i < 10:
driver.get("http://www.example.com")
txt = driver.find_element_by_id('input-text')
txt.clear()
txt.send_keys(msg)
btn = driver.find_element_by_id('input-search')
btn.click()
i += 1
driver.quit()
I think it's because in my test, there are two things:
1) fill abc in input-text;
2) click a button, submit the abc and open a new web page.
On the new page, there is also an input-text and a button input-search, so it will fill the abc and click the button on the new page, which is not what I want.

At btn.click(), it doesn't wait for page loaded because it is not a hyperlink and it doesn't submit a webform. So your script might be failed. You should fix it by waiting for some elements to determine if it should reload page or not. See code below.
try:
driver.get('http://www.example.com')
txt = WebDriverWait(driver, 10).until(EC.presence_of_element_located((By.ID, "input-text")))
txt.clear()
txt.send_keys(msg)
btn = driver.find_element_by_id('input-search')
btn.click()
countryDescriptionElement = WebDriverWait(driver, 10).until(EC.presence_of_element_located((By.ID, "country-description")))
#print(driver.page_source.encode('utf-8'))
except WebDriverException as ex:
print("Enter: " + msg + ", Error: " + str(ex) + ", Found: " + driver.page_source)

Unable to click element on page Selenium python

I am trying to move to page 2 and beyond of this page (pagination) with python selenium and spent a few hours on this. I am getting this error, and would be thankful of any help..Error from chromedriver
is not clickable at point(). Other element would receive the click
My code so far:
class Chezacash:
t1 = time.time()
driver = webdriver.Chrome(chromedriver)
def controller(self):
self.driver.get("https://www.chezacash.com/#/home/")
element = WebDriverWait(self.driver, 10).until(
EC.presence_of_element_located((By.CSS_SELECTOR, "div.panel-heading")))
soup = BeautifulSoup(self.driver.page_source.encode('utf-8'),"html.parser")
self.parser(soup)
self.driver.find_element(By.XPATH, "//li[#class='paginate_button active']/following-sibling::li").click()
time.sleep(2)
soup = BeautifulSoup(self.driver.page_source.encode('utf-8'),"html.parser")
self.parser(soup)
def parser(self, soup):
for i in soup.find("table", {"id":"DataTables_Table_1"}).tbody.contents:
date = i.findAll("td")[0].get_text().strip()
time = i.findAll("td")[1].get_text().strip()
home = i.findAll("td")[4].div.span.get_text().strip().encode("utf-8")
home_odds = i.findAll("td")[4].div.findAll("span")[1].get_text().strip()
draw_odds = i.findAll("td")[5].div.findAll("span")[1].get_text().strip()
away = i.findAll("td")[6].div.span.get_text().strip().encode("utf-8")
away_odds = i.findAll("td")[6].div.findAll("span")[1].get_text().strip()
print home
cheza = Chezacash()
try:
cheza.controller()
except:
cheza.driver.service.process.send_signal(signal.SIGTERM) # kill the specific phantomjs child proc # quit the node proc
cheza.driver.quit()
traceback.print_exc()

What if instead you would locate the "Next" button by link text, scroll into it's view and then click:
next_button = self.driver.find_element_by_link_text("Next")
self.driver.execute_script("arguments[0].scrollIntoView();", next_button)
next_button.click()
I would also maximize the browser window before navigating to the page:
self.driver.maximize_window()
self.driver.get("https://www.chezacash.com/#/home/")