Develop Reference

Getting text from a div + code object with Selenium

I'm trying to get the text inside these div's but I'm not succeeding. They have a code between the class that doesn't change with each execution.
<div data-v-a5b90146="" class="html-content"> Text to be captured</div>
<div data-v-a5b90146="" class="html-content"><b> TEXT WANTED </b><div><br></div>
I've tried with XPATH, but I was not successful too.
content = WebDriverWait(driver, 10).until(EC.presence_of_all_elements_located((By.XPATH, '/html/body/div/div/div/div/div/div/div[1]/div[2]/div[2]/div[4]/div/div/b'))).text

You need to change couple of things.
presence_of_all_elements_located() returns list of elements, so you can't use .text with a list. To get the text value of the element you need to iterate and then get the text.
xpath seems very fragile. you should use relative xpath since class name is unique you can use the classname.
your code should be like.
contents = WebDriverWait(driver, 10).until(EC.presence_of_all_elements_located((By.XPATH, '//div[#class="html-content"][#data-v-a5b90146]')))
for content in contents:
print(content.text)
You can use visibility_of_all_elements_located() as well
contents = WebDriverWait(driver, 10).until(EC.visibility_of_all_elements_located((By.XPATH, '//div[#class="html-content"][#data-v-a5b90146]')))
for content in contents:
print(content.text)

Both the <div> tags have the attribute class="html-content"
Solution
To extract the texts from the <div> tags instead of presence_of_all_elements_located(), you have to induce WebDriverWait for visibility_of_all_elements_located() and using list comprehension and you can use either of the following locator strategies:
Using CSS_SELECTOR:
print([my_elem.text for my_elem in WebDriverWait(driver, 20).until(EC.visibility_of_all_elements_located((By.CSS_SELECTOR, "div.html-content")))])
Using XPATH:
print([my_elem.text for my_elem in WebDriverWait(driver, 20).until(EC.visibility_of_all_elements_located((By.XPATH, "//div[#class='html-content']")))])
Note : You have to add the following imports :
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.common.by import By
from selenium.webdriver.support import expected_conditions as EC

Try to get the element first then get the text from the element;
element = WebDriverWait(
driver,
10
).until(
EC.presence_of_all_elements_located((
By.XPATH, '/html/body/div/div/div/div/div/div/div[1]/div[2]/div[2]/div[4]/div/div/b'
))
)
content = element.text

Getting list of all the URLs in a Closed Issue page in GitHub using Selenium

I am trying to store the links of all the closed issues from a GitHub (https://github.com/mlpack/mlpack/issues?q=is%3Aissue+is%3Aclosed) project using Selenium. I use the code below:
repo_closed_url = [link.find_element(By.CLASS_NAME,'h4').get_attribute('href') for link in driver.find_elements(By.XPATH,'//div[#aria-label="Issues"]')]
However, the above code only returns the first URL. How can I get all the URLs in that page? I iterate through all the pages. So just getting the links from the first page is fine.

Please try this. This should work:
repo_closed_url = [link.get_attribute('href') for link in driver.find_elements(By.XPATH,"//div[#aria-label='Issues']//a[contains(#class,'h4')]")]
Here //div[#aria-label='Issues']//a[contains(#class,'h4')] XPath locates directly all the desired title elements on the page.
Then the rest of the code in the line is iterating over the list of returning elements extracting their href attributes as I explained in the previous question.

To extract the links from all the href attributes you have to induce WebDriverWait for visibility_of_all_elements_located() and you can use either of the following locator strategies:
Using CSS_SELECTOR:
driver.get("https://github.com/mlpack/mlpack/issues?q=is%3Aissue+is%3Aclosed")
print([my_elem.get_attribute("href") for my_elem in WebDriverWait(driver, 20).until(EC.visibility_of_all_elements_located((By.CSS_SELECTOR, "div[id^='issue_'] a[id^='issue']")))])
Using XPATH:
driver.get("https://github.com/mlpack/mlpack/issues?q=is%3Aissue+is%3Aclosed")
print([my_elem.get_attribute("href") for my_elem in WebDriverWait(driver, 20).until(EC.visibility_of_all_elements_located((By.XPATH, "//div[starts-with(#id, 'issue_')]//a[starts-with(#id, 'issue')]")))])
Console Output:
['https://github.com/mlpack/mlpack/issues/3371', 'https://github.com/mlpack/mlpack/issues/3370', 'https://github.com/mlpack/mlpack/issues/3369', 'https://github.com/mlpack/mlpack/issues/3368', 'https://github.com/mlpack/mlpack/issues/3367', 'https://github.com/mlpack/mlpack/issues/3365', 'https://github.com/mlpack/mlpack/issues/3364', 'https://github.com/mlpack/mlpack/issues/3363', 'https://github.com/mlpack/mlpack/issues/3356', 'https://github.com/mlpack/mlpack/issues/3353', 'https://github.com/mlpack/mlpack/issues/3352', 'https://github.com/mlpack/mlpack/issues/3351', 'https://github.com/mlpack/mlpack/issues/3348', 'https://github.com/mlpack/mlpack/issues/3340', 'https://github.com/mlpack/mlpack/issues/3338', 'https://github.com/mlpack/mlpack/issues/3336', 'https://github.com/mlpack/mlpack/issues/3333', 'https://github.com/mlpack/mlpack/issues/3329', 'https://github.com/mlpack/mlpack/issues/3326', 'https://github.com/mlpack/mlpack/issues/3325', 'https://github.com/mlpack/mlpack/issues/3324', 'https://github.com/mlpack/mlpack/issues/3323', 'https://github.com/mlpack/mlpack/issues/3319', 'https://github.com/mlpack/mlpack/issues/3314', 'https://github.com/mlpack/mlpack/issues/3303']
Note : You have to add the following imports :
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.common.by import By
from selenium.webdriver.support import expected_conditions as EC

Try the below XPath expression:
//div[#aria-label='Issues']//a[contains(#id,'issue')]
This XPath expression will list all the closed issues in page 1. Just use .get_attribute('href') to get the URLs.

Selenium Python: Flagging pages with specific span text

I currently have a Selenium script that is running through a list of part numbers on a website and capturing some information such as product name (pulled from the page title).
I have noticed that some of the product is identified as "DISCONTINUED" (through a span) and so I would like to be able to capture that information so that I can ignore all of those products.
On the website in general, they denote these products through:
<span data*="">DISCONTINUED</span>
Any other products that are valid will not have this information on it, and so in this case I want to ensure that the script doesn't crash and just captures a blank value.
I tried using:
driver.find_element(By.XPATH, '//html/body/div/div/section/div/section/div/div/section/div/div/div/div/div/span/span[text()="DISCONTINUED"]')
However I get this error:
selenium.common.exceptions.NoSuchElementException: Message: no such element: Unable to locate element: {"method":"xpath","selector":"//html/body/div/div/section/div/section/div/div/section/div/div/div/div/div/span/span[text()="DISCONTINUED"]"}
I did make sure to use only a single part number that did indeed have a DISCONTINUED on it.
I also did load up the page in dev tools and searched for that specific xpath and it did indeed highlight the proper section:
Searched:
//html/body/div/div/section/div/section/div/div/section/div/div/div/div/div/span/span
Highlighted:
<span data*="">DISCONTINUED</span>
What is the best way to do this? Also, I need to check to see how to include a blank value by default so that it will not crash when a current product is retrieved.

Considering the HTML:
<span data*="">DISCONTINUED</span>
To locate the element with the text DISCONTINUED you can use either of the following locator strategies:
Using xpath and text():
element = driver.find_element(By.XPATH, "//span[text()='DISCONTINUED']")
Using xpath and contains():
element = driver.find_element(By.XPATH, "//span[contains(., 'DISCONTINUED')]")
Ideally to locate the element you need to induce WebDriverWait for the visibility_of_element_located() and you can use either of the following locator strategies:
Using xpath and text():
element = WebDriverWait(driver, 20).until(EC.visibility_of_element_located((By.XPATH, "//span[text()='DISCONTINUED']")))
Using xpath and contains():
element = WebDriverWait(driver, 20).until(EC.visibility_of_element_located((By.XPATH, "//span[contains(., 'DISCONTINUED')]")))
Note : You have to add the following imports :
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.common.by import By
from selenium.webdriver.support import expected_conditions as EC

This is what I used to make it work with error avoidance.
try:
discontinued = WebDriverWait(driver, 2).until(EC.visibility_of_element_located((By.XPATH, "//html/body/div/div/section/div/section/div/div/section/div/div/div/div/div/span/span[text()='ARCHIVED']")))
discontinued = "CURRENT"
except:
discontinued="DISCONTINUED"
Thanks again for the guidance on using waits instead of the find_element approach!

Selenium Python: Element is not clickable

I am testing a webapp created with python dash using selenium. I am trying to click a tab but always get the ElementClickIntercepted Exception.
Note: Of course I stumbled over similiar problems but most times the problem is that the element cannot be clicked as is did not load yet - I already built in waits! - or the fact that another element would receive the click which cannot be the case either. Note that the first item is always preselected and i want to choose the second.
#Selct X-Axis
x_axis = driver.find_element(By.XPATH, "//*[#id='navbar']/a[2]")
x_axis.click()
driver.implicitly_wait(2)
try:
WebDriverWait(driver, 10).until(
EC.presence_of_element_located((By.XPATH, "//a[text()='Even']"))
)
except:
raise Exception
even= driver.find_element(By.XPATH, "//a[text()='Even']")
even.click()
Exception snapshot:
HTML snapshot:

Based on what you've posted, By.XPATH, "//a[text()='Even']" is probably selecting an HTML element different from what you're expecting.
There appears to be an <a> with text that starts with 'Even', but you've shown nothing with text that equals 'Even'.
BTW, when asking a question, you ought to at least paste the relevant HTML as text, rather than as a screenshot.

Use contains(text(), 'even') instead of text()='even' in your XPath. Because, text()='even' means the text must be 'even' but not contain 'even', and in your HTML code, the text seems not only have 'even'.

To click on either of the element instead of presence_of_element_located() you need to induce WebDriverWait for the element_to_be_clickable() and you can use either of the following locator strategies:
$$$ Response:
Using PARTIAL_LINK_TEXT:
WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.PARTIAL_LINK_TEXT, "Response"))).click()
Using XPATH:
WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.XPATH, "//div[starts-with(#id, 'content')]//ul//div[#class='nav-item']//a[contains(., 'Response')]"))).click()
Even $$$:
Using PARTIAL_LINK_TEXT:
WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.PARTIAL_LINK_TEXT, "Even"))).click()
Using XPATH:
WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.XPATH, "//div[starts-with(#id, 'content')]//ul//div[#class='nav-item']//a[starts-with(., 'Even')]"))).click()
Note: You have to add the following imports :
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.common.by import By
from selenium.webdriver.support import expected_conditions as EC

Python Selenium - get href value

I am trying to copy the href value from a website, and the html code looks like this:
<p class="sc-eYdvao kvdWiq">
<a href="https://www.iproperty.com.my/property/setia-eco-park/sale-
1653165/">Shah Alam Setia Eco Park, Setia Eco Park
</a>
</p>
I've tried driver.find_elements_by_css_selector(".sc-eYdvao.kvdWiq").get_attribute("href") but it returned 'list' object has no attribute 'get_attribute'. Using driver.find_element_by_css_selector(".sc-eYdvao.kvdWiq").get_attribute("href") returned None. But i cant use xpath because the website has like 20+ href which i need to copy all. Using xpath would only copy one.
If it helps, all the 20+ href are categorised under the same class which is sc-eYdvao kvdWiq.
Ultimately i would want to copy all the 20+ href and export them out to a csv file.
Appreciate any help possible.

You want driver.find_elements if more than one element. This will return a list. For the css selector you want to ensure you are selecting for those classes that have a child href
elems = driver.find_elements_by_css_selector(".sc-eYdvao.kvdWiq [href]")
links = [elem.get_attribute('href') for elem in elems]
You might also need a wait condition for presence of all elements located by css selector.
elems = WebDriverWait(driver,10).until(EC.presence_of_all_elements_located((By.CSS_SELECTOR, ".sc-eYdvao.kvdWiq [href]")))

As per the given HTML:
<p class="sc-eYdvao kvdWiq">
Shah Alam Setia Eco Park, Setia Eco Park
</p>
As the href attribute is within the <a> tag ideally you need to move deeper till the <a> node. So to extract the value of the href attribute you can use either of the following Locator Strategies:
Using css_selector:
print(driver.find_element_by_css_selector("p.sc-eYdvao.kvdWiq > a").get_attribute('href'))
Using xpath:
print(driver.find_element_by_xpath("//p[#class='sc-eYdvao kvdWiq']/a").get_attribute('href'))
If you want to extract all the values of the href attribute you need to use find_elements* instead:
Using css_selector:
print([my_elem.get_attribute("href") for my_elem in driver.find_elements_by_css_selector("p.sc-eYdvao.kvdWiq > a")])
Using xpath:
print([my_elem.get_attribute("href") for my_elem in driver.find_elements_by_xpath("//p[#class='sc-eYdvao kvdWiq']/a")])
Dynamic elements
However, if you observe the values of class attributes i.e. sc-eYdvao and kvdWiq ideally those are dynamic values. So to extract the href attribute you have to induce WebDriverWait for the visibility_of_element_located() and you can use either of the following Locator Strategies:
Using CSS_SELECTOR:
print(WebDriverWait(driver, 10).until(EC.visibility_of_element_located((By.CSS_SELECTOR, "p.sc-eYdvao.kvdWiq > a"))).get_attribute('href'))
Using XPATH:
print(WebDriverWait(driver, 10).until(EC.visibility_of_element_located((By.XPATH, "//p[#class='sc-eYdvao kvdWiq']/a"))).get_attribute('href'))
If you want to extract all the values of the href attribute you can use visibility_of_all_elements_located() instead:
Using CSS_SELECTOR:
print([my_elem.get_attribute("innerHTML") for my_elem in WebDriverWait(driver, 20).until(EC.visibility_of_all_elements_located((By.CSS_SELECTOR, "p.sc-eYdvao.kvdWiq > a")))])
Using XPATH:
print([my_elem.get_attribute("innerHTML") for my_elem in WebDriverWait(driver, 20).until(EC.visibility_of_all_elements_located((By.XPATH, "//p[#class='sc-eYdvao kvdWiq']/a")))])
Note : You have to add the following imports :
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.common.by import By
from selenium.webdriver.support import expected_conditions as EC

The XPATH
//p[#class='sc-eYdvao kvdWiq']/a
return the elements you are looking for.
Writing the data to CSV file is not related to the scraping challenge. Just try to look at examples and you will be able to do it.

To crawl any hyperlink or Href, proxycrwal API is ideal as it uses pre-built functions for fetching desired information. Just pip install the API and follow the code to get the required output. The second approach to fetch Href links using python selenium is to run the following code.
Source Code:
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
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.webdriver.support.ui import Select
import time
list = ['https://www.heliosholland.com/Ampullendoos-voor-63-ampullen','https://www.heliosholland.com/lege-testdozen’]
driver = webdriver.Chrome()
wait = WebDriverWait(driver,29)
for i in list:
driver.get(i)
image = wait.until(EC.visibility_of_element_located((By.XPATH,'/html/body/div[1]/div[3]/div[2]/div/div[2]/div/div/form/div[1]/div[1]/div/div/div/div[1]/div/img'))).get_attribute('src')
print(image)
To scrape the link, use .get_attribute(‘src’).

Get the whole element you want with driver.find_elements(By.XPATH, 'path').
To extract the href link use get_attribute('href').
Which gives,
driver.find_elements(By.XPATH, 'path').get_attribute('href')

try something like:
elems = driver.find_elements_by_xpath("//p[contains(#class, 'sc-eYdvao') and contains(#class='kvdWiq')]/a")
for elem in elems:
print elem.get_attribute['href']