How can I fill the form of this popup? - python

I'm just learning Selenium and I wrote this small code. From there, I don't find any way to fill a form. The HTML source of the fill popups appears only when you clic "Créer un compte", "Create an account" in english.
But it looks like when I just print(driver.source_page) I don't get this part of the code.
from selenium import webdriver
from selenium.webdriver.common.by import By
import webbrowser
driver = webdriver.Chrome()
driver.get('http://labrute.muxxu.com/')
element = driver.find_element(By.CLASS_NAME,'tid_long')
element.click()
driver.switch_to_frame(driver.find_element_by_css_selector('//*[#id="create"]/div[1]/div[1]/div[1]/div/label'))
Do you have any ideas?
Here is the popup
Here is without popup
Thanks
I tried to use like driver.switch_to xx methods, but nothing worked yet.

It is inside an iframe, you have to switch to the iframe then you have to enter the values:
WebDriverWait(driver, 15).until(EC.frame_to_be_available_and_switch_to_it((By.XPATH, "//*[#name='tid_create']")))
time.sleep(1)
driver.find_element(By.XPATH, "//*[#name='name']").send_keys("name")

Thanks for your help,
The code finally looks like:
driver.switch_to.frame(driver.find_element(By.CSS_SELECTOR,'#tid_frameWrapper
> iframe'))
Best regards

Related

How to make nonclickable button click able in python using selenium?

Hi Before starting Thanks for the help in advance
So I am trying to scrape google flight website : https://www.google.com/travel/flights
When scraping I have done the sending Key to the text field but I am stuck at clicking the search button it always gives the error that the field is not clickable at a point or Other elements would receive the click
the error image is
and the code is
from time import sleep
from selenium import webdriver
chromedriver_path = 'E:/chromedriver.exe'
def search(urrl):
driver = webdriver.Chrome(executable_path=chromedriver_path)
driver.get(urrl)
asd= "//div[#aria-label='Enter your destination']//div//input[#aria-label='Where else?']"
driver.find_element_by_xpath("/html/body/c-wiz[2]/div/div[2]/c-wiz/div/c-wiz/c-wiz/div[2]/div[1]/div[1]/div[1]/div[2]/div[1]/div[4]/div/div/div[1]/div/div/input").click()
sleep(2)
TextBox = driver.find_element_by_xpath(asd)
sleep(2)
TextBox.click()
sleep(2)
print(str(TextBox))
TextBox.send_keys('Karachi')
sleep(2)
search_button = driver.find_element_by_xpath('//*[#id="yDmH0d"]/c-wiz[2]/div/div[2]/c-wiz/div/c-wiz/c-wiz/div[2]/div[1]/div[1]/div[2]/div/button/div[1]')
sleep(2)
search_button.click()
print(str(search_button))
sleep(15)
print("DONE")
driver.close()
def main():
url = "https://www.google.com/travel/flights"
print(" Getitng Data ")
search(url)
if __name__ == "__main__":
main()
and i have done it by copying the Xpath using dev tools
Thanks again
The problem you are facing is that after entering the city Karachi in the text box, there is a suggestion dropdown that is displayed over the Search Button. That is the cause of the exception as the dropdown would receive the click instead of the Search button. The intended usage of the website is to select the city from the dropdown and then continue.
A quick fix would be to first look for all of the dropdowns in the source (there are a few) and look for the one that is currently active using is_displayed(). Next you would select the first element in the dropdown suggested:
.....
TextBox.send_keys('Karachi')
sleep(2)
# the attribute(role) in the dropdowns element are named 'listbox'. Warning: This could change in the future
all_dropdowns = driver.find_elements_by_css_selector('ul[role="listbox"]')
# the active dropdown
active_dropdown = [i for i in all_dropdowns if i.is_displayed()][0]
# click the first suggestion in the dropdown (Note: this is very specific to your search. It could not be desirable in other circumstances and the logic can be modified accordingly)
active_dropdown.find_element_by_tag_name('li').click()
# I recommend using the advise #cruisepandey has offered above regarding usage of relative path instead of absolute xpath
search_button = driver.find_element_by_xpath("//button[contains(.,'Search')]")
sleep(2)
search_button.click()
.....
Also suggest to head the advise provided by #cruisepandey including research more about Explicit Waits in selenium to write better performing selenium programs. All the best
Instead of this absolute xapth
//*[#id="yDmH0d"]/c-wiz[2]/div/div[2]/c-wiz/div/c-wiz/c-wiz/div[2]/div[1]/div[1]/div[2]/div/button/div[1]
I would recommend you to have a relative path:
//button[contains(.,'Search')]
Also, I would recommend you to have explicit wait when you are trying a click operation.
Code:
search_button = WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.XPATH, "//button[contains(.,'Search')]")))
search_button.click()
You'd need below imports as well:
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.common.by import By
from selenium.webdriver.support import expected_conditions as EC
Pro tip:
Launch the browser in full screen mode.
driver.maximize_window()
You should have the above line just before driver.get(urrl) command.

How to pass the RGPD popup / iframe with Selenium (Python)?

Ok, attempt to use Selenium on a website : lefigaro.fr, but no class related to the RGPD popup to be found by Selenium, even after a switch to frame. :/
I'm juste looking a reliable way to close it.
It goes this way :
from selenium import webdriver
WINDOW_SIZE = "1920,1080"
ADRESSE = 'https://www.lefigaro.fr/'
driver = webdriver.Firefox() #the chrome version was even worse, geckodriver a the root
driver.get(ADRESSE)
elt = driver.find_element_by_class_name("sc-18sn7k8-1") #error
driver.switch_to.frame(driver.find_element_by_tag_name("iframe"))
elt = driver.find_element_by_class_name("sc-18sn7k8-1") #error
You can switch to iframe via index.
It seems that popup is the only iframe on the page, so perhaps this will work:
driver.switch_to.frame(0) # or perhaps (1)
EDIT:
It seems the page already is in an iframe when it loads.
Do driver.switch_to.default_content() first.
Then it seems the iframe you want is (3):
driver.switch_to.frame(3)
Hope this works.
Hmm, actually, not quite... I'll spend some more time on this.

Click a button using Selenium on python

I am trying to click this button to export my data as a csv file. I'm using Selenium on python.
I can not figure out how to press the button because there is no id or name and the link changes every time.
Thanks in advance!
You have to scroll to that element. Like this:
from selenium.webdriver.common.action_chains import ActionChains
element = driver.find_element_by_xpath("//a[contains(#href,'swissadme.csv')]")
actions = ActionChains(driver)
actions.move_to_element(element).perform()
element.click()
If the locator above not enough try this instead:
"//a[contains(#href,'results/50881847/swissadme.csv')]"
See if this works:-
elm = driver.find_element_by_xpath(".//div[contains(text(),'Retrieve date:')]/a[contains(#href,''swissadme.csv)]")
driver.execute_script("arguments[0].scrollIntoView(true);",elm);
elm.click()

Cannot click on an xpath selected object Selenium (Python)

I am trying to click to an object that I select with Xpath, but there seems to be problem that I could not located the element. I am trying to click accept on the page's "Terms of Use" button. The code I have written is as
driver.get(link)
accept_button = driver.find_element_by_xpath('//*[#id="terms-ok"]')
accept_button.click()
prov = driver.find_element_by_id("province-region")
prov.click()
Here is the HTML code I have:
And I am getting a "NoSuchElementException". My goal is to click this "Kabul Ediyorum" button at the bottom of the HTML code. I started to think that we have some restrictions on the tasks we could do on that page. Any ideas ?
Not really sure what the issue might be.
But you could try the following:
Try to locate the element by its visible text
accept_button = driver.find_element_by_xpath("//*[text()='Kabul Ediyorum']").click()
Try with ActionChains
For that you need to import ActionChains
from selenium.webdriver.common.action_chains import ActionChains
accept_button = driver.find_element_by_xpath("//*[text()='Kabul Ediyorum']")
actions = ActionChains(driver)
actions.click(on_element=accept_button).perform()
Also make sure you have implicitly wait
# implicitly wait
driver.implicitly_wait(10)
Or explicit wait
element = WebDriverWait(driver, 10).until(EC.element_to_be_clickable((By.XPATH, "//*[text()='Kabul Ediyorum']"))).click()
Hope this helped!

Give upload file path to Instagram with Selenium and python

I'm testing some web scraping on Instagram with Selenium and Python.
In this case I want to upload a picture.
Normally you have to click on the upload icon and choose the file from a window. How can I manage it with Selenium?
I tried:
driver.find_element_by_class_name("coreSpriteFeedCreation").send_keys('C:\\path-to-file\\file.jpg')
and also with find_element_by_xpath but I get an exception:
selenium.common.exceptions.WebDriverException: Message: unknown error: cannot focus element
I tried also only with click() but nothing happens.
Any Idea?
EDIT
Thanks to #homersimpson comment I tried this:
actions = ActionChains(driver)
element = driver.find_element_by_class_name("coreSpriteFeedCreation")
actions.move_to_element(element)
actions.click()
actions.send_keys('C:\\path-to-file\\file.jpg')
actions.perform()
Now the window to choose the file appears. The problem is that I would like to avoid this window and give directly the path of my file.
If right understand, you are trying to avoid handling with a native window. You can try this:
# get all inputs
inputs = driver.find_elements_by_xpath("//input[#accept = 'image/jpeg']").send_keys(os.getcwd() + "/image.png")
Now you can try all of them. I don't know which of them will work.
More about os.getcwd() is here
To be able to perform this code you have to have an element like this:
<input type="file" name="fileToUpload" id="fileToUpload2" class="fileToUpload">
EDIT:
It looks like instagram turned of input fields interaction for posts. For Account image it still works, but not for posting. I assume it is was done to prevent bots to post images. Anyway, there is a solution for this problem. You can use AutoIt like this:
import autoit
from selenium import webdriver
from selenium.webdriver.common.action_chains import ActionChains
from selenium.webdriver.common.desired_capabilities import DesiredCapabilities
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.webdriver.common.keys import Keys
ActionChains(driver).move_to_element( driver.find_element_by_xpath("//path/to/upload/button")).click().perform()
handle = "[CLASS:#32770; TITLE:Open]"
autoit.win_wait(handle, 60)
autoit.control_set_text(handle, "Edit1", "\\file\\path")
autoit.control_click(handle, "Button1")
I think I may have found a solution that works for me. I found if you first have the bot click the plus icon while the browser is in the mobile view.
self.driver.find_element_by_xpath("/html/body/div[1]/section/nav[2]/div/div/div[2]/div/div/div[3]")\
.click()
after that, I would immediately send my file to an input tag in the HTML and I find you may need to play around as to which one works but I find the last input tag worked for me.
self.driver.find_element_by_xpath("/html/body/div[1]/section/nav[2]/div/div/form/input")\
.send_keys("/image/path.jpg")
The one weird thing about this is you will have a popup menu on top of the page but your code will still function with this window displayed over the window you are working on.
Addition to HumbleFox's answer. To solve his problem regarding the pop-up box not closing or the file pop-up box not closing (bug)
The solution to this is to make the browser headless here's a part of my code for example:
mobile_emulation = { "deviceName": "Pixel 2" }
chrome_options = webdriver.ChromeOptions()
chrome_options.add_argument('--headless')
chrome_options.add_experimental_option("mobileEmulation", mobile_emulation)
chrome_options.binary_location = self.opt.binary_location
self.driver = webdriver.Chrome(executable_path=self.path, options=chrome_options)

Categories