Can't Find Javascript href Link in Python Webscrape

Question

I'm trying to webscrape this site: https://www2.tse.or.jp/tseHpFront/JJK020010Action.do

Using the Selenium package, with Google Chrome as my browser, I'm able to open it up, choose some settings, and then run a search. I'm encountering an error because there are 21 pages of information, and I need to gather all of it, however with my code I am unable to find the link that clicks to the next page. This is what that next button's code is:

<div class="next_e">
   <a href="javascript:setPage(2);submitPage(document.JJK020030Form, document.JJK020030Form.Transition);">
      <img src="/common/images/spacer.gif"  width="77"  height="24"  alt="Next">
   </a>
</div>

Note -- the number in the brackets after 'setPage' corresponds to the next page number. So if I'm on page 1 the code would read setPage(2), etc.

Here is my complete code for the webscrape:

driver.get("https://www2.tse.or.jp/tseHpFront/JJK020030Action.do")
sleep(20)
data = []

button = driver.find_element_by_name("dspSsuPd")
#driver.find_elements_by_class_name
button200 = Select(button)
button200.select_by_value('200')

sleep(10)

checkboxes = ['001', '002', '004', '006', '008', '101', '102', '104', 'ETF', 'ETN', 'RET', 'PSC', '999']
for box in checkboxes:
    driver.find_element_by_xpath(f"//input[@value='{box}']").click()

search_button = "//*[@class='activeButton' and @value='Start of search']"
driver.find_element(By.XPATH, search_button).click()
sleep(20)

soup1 = BeautifulSoup(driver.page_source, 'lxml')
tables1 = soup.find_all('table')
df = pd.read_html(driver.page_source)[-1]
data.append(df)

for i in range(2, 21):
    
## right here is where I'm encountering my issue ##
    next_href = f"//*[@class='next_e' and @href ='javascript:setPage({i});submitPage(document.JJK020030Form, document.JJK020030Form.Transition);']"
    driver.find_element(By.XPATH, next_href).click()
    sleep(10)

    soup = BeautifulSoup(driver.page_source, 'lxml')
    tables = soup.find_all('table')
    df1 = pd.read_html(driver.page_source)[-1]
    data.append(df1)

driver.quit()
df_data = pd.DataFrame(pd.concat(data)).reset_index(drop=True)
print(df_data)
df_data.to_csv('companies_data_borse_frankfurt.csv', index=False)

The other options I have tried to click this href (all of which haven't worked), include:

driver.find_element(By.XPATH, "//div[@class='next_e']/a[contains(., 'setPage')]").click()

WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.XPATH, "//div[@class='next_e']/a[contains(., 'setPage')]"))).click()

driver.find_element_by_xpath(f'//input[@href="javascript:setPage({i});submitPage(document.JJK020030Form, document.JJK020030Form.Transition);"]').click()

driver.find_element_by_partial_link_text(f'javascript:setPage({i})')

Please let me know if you have a solution or need further clarification on the issue. Thank you!

Arundeep Chohan · Accepted Answer · 2022-01-31 03:31:05Z

1

wait=WebDriverWait(driver,60)      
wait.until(EC.element_to_be_clickable((By.CSS_SELECTOR,"div.next_e>a"))).click()

Using this work just fine for going through the pages.

Import:

from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait 
from selenium.webdriver.support import expected_conditions as EC

answered Jan 31, 2022 at 3:31

Arundeep Chohan

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2 Comments

Arundeep Chohan Over a year ago

You could trim some code off since starting on page and then incrementing would just be better here.

kylala Over a year ago

Unfortunately I can't. This site doesn't work that way :/

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