Develop Reference

How do I click on an item on my google search page with Selenium Python?

Good time of the day!
Faced with a seemingly simple problem,
But it’s been a while, and I’m asking for your help.
I work with Selenium on Python and I need to curse about
20 items on google search page by random request.
And I’ll give you an example of the elements below, and the bottom line is, once the elements are revealed,
Google generates new such elements
Problem:
Cannot click on the element. I will need to click on existing and then on new, generated elements in this block: click for open see blocks with element
Tried to click on xpath, having collected all the elements:
xpath = '//*[#id="qmCCY_adG4Sj3QP025p4__16"]/div/div/div[1]/div[4]'
all_elements = driver.find_element(By.XPATH, value=xpath)
for element in all_elements:
element.click()
sleep(2)
Important note!
id xpath has constantly changing and is generated by another on the google side
Tried to click on the class
class="r21Kzd"
Tried to click on the selector:
#qmCCY_adG4Sj3QP025p4__16 > div > div > div > div.wWOJcd > div.r21Kzd
Errors
This is when I try to click using xpath:
Message: no such element: Unable to locate element: {"method":"xpath","selector"://*[#id="vU-CY7u3C8PIrgTuuJH4CQ_9"]/div/div[1]/div[4]}
In other cases, the story is almost the same, the driver does not find the element and cannot click on it. Below I apply a scratch tag on which I need to click
screenshot tags on google search
Thanks for the help!

In case iDjcJe IX9Lgd wwB5gf are a fixed class name values of that element all you need is to use CSS_SELECTOR instead of CLASS_NAME with a correct syntax of CSS Selectors.
So, instead of driver.find_element(By.CLASS_NAME, "iDjcJe IX9Lgd wwB5gf") try using this:
driver.find_element(By.CSS_SELECTOR, ".iDjcJe.IX9Lgd.wwB5gf")
(dots before each class name, no spaces between them)

How do I click a button using Selenium in Python?

I'm trying to crawl review data on an app from a Google Play website page using Python and Selenium. What I'm trying to do here is to click the "Full Review" button to see the entire text of each review.
But I keep running into these errors each time:
ElementNotInteractableException: Message: element not interactable
or
AttributeError: 'list' object has no attribute 'click'
or
NoSuchElementException: Message: no such element: Unable to locate element: {"method":"css selector","selector":".LkLjZd ScJHi OzU4dc "}
The element for the button is this:
<button class="LkLjZd ScJHi OzU4dc " jsaction="click:TiglPc" jsname="gxjVle">Full Review</button>
The xpath for the button is this:
//*[#id="fcxH9b"]/div[4]/c-wiz/div/div[2]/div/div[1]/div/div/div[1]/div[2]/div/div[26]/div/div[2]/div[2]/span[1]/div/button
And this is the code I'm using for xpath:
full_review = driver.find_elements_by_xpath('//*[#id="fcxH9b"]/div[4]/c-wiz/div/div[2]/div/div[1]/div/div/div[1]/div[2]/div/div[26]/div/div[2]/div[2]/span[1]/div/button')
full_review.click()
I can't find the button by class name, xpath or anything.
Could anyone help me figure out this problem? In addition, is it possible to find the element by jsname?
I would really appreciate if you could help.

Avoid using xpath whenever possible, it's the most brittle selector.
id > CSS > Xpath
For your button, this css selector should work:
button[jsname='gxjVle']
You'll probably need to specify the child as that probably won't be unique

Your XPath selector is a little bit flaky, you can simplify it to be something like:
//button[text()='Full Review']
You're using the wrong function, if you're looking for a single unique element you should be using WebDriver.find_element_by_xpath one, mind that element, not elements
It's better to use Explicit Wait just in case element is not immediately available in DOM (for example if you're testing application built using AJAX technology)
Assuming all above:
full_review = WebDriverWait(driver, 10).until(EC.element_to_be_clickable((By.XPATH, "//button[text()='Full Review']")))

How to find href through class name and click with selenium

I'm using Selenium with python to make a spider.
A part of the web page is like this:
text - <span class="sr-keyword">name</span> text
I need to find the href and click.
I've tried as below:
target = driver.find_element_by_class_name('_j_search_link')
target.click()
target is not None but it doesn't seem that it could be clicked because after target.click(), nothing happened.
Also, I've read this question: Click on hyperlink using Selenium Webdriver
But it can't help me because in my case, there is a <span class>, not just a simple text Google.

You can look for an element with class _j_search_link that contains text Wat Chedi Luang
driver.find_element_by_xpath('//a[#class="_j_search_link" and contains(., "Wat Chedi Luang")]')

driver.find_elements_by_partial_link_text("Wat Chedi Luang").click()

I think you didn't find right element. Use CSS or Xpath to target this element then click it, like:
//a[#class='_j_search_link']
or
//a[#class='_j_search_link']/span[#class='sr-keyword']

Unable to locate element python selenium - Need to click this button

So, this button has no id/name and I've tried xpath, class name, CSS selector, etc
Here is the HTML code:
Thanks in advance.

Be sure to switchTo the iframe that contains this button! Then I think that this CSS path should work:
div.submit-and-reset-wrapper button.continue

You can try a kind of ugly hack:
get all the button/span elements
loop all the button/span elements until you find the text.
then click the element...
something like:
all_span_elements = driver.find_elements_by_tag_name("span")
for span_element in all_span_elements:
if "Begin Lesson" in span_element.text:
span_element.click()

How to send keys to a difficult element in selenium (Python)

I am trying to click a small button, which has no "ID" or "Name", on a website. The only unique identifier is onclick, which is as follows:
onclick="submitForm('DefaultFormName',1,{'param1':'HXCTIMECARDACTIVITIESPAGEXgm7J5oT','serverValidate':'1uCdqvhJe','param2':'',event:'details|1'});return false;"
However, another button on the page has the following onclick:
onclick="submitForm('DefaultFormName',1,{'param1':'HXCTIMECARDACTIVITIESPAGEXgm7J5oT','serverValidate':'1uCdqvhJe','param2':'',event:'details|2'});return false;"
The only difference is a 1 vs. a 2 in the middle of the code. I tried to use a "find_element_by_css_selector" with the following code but it didn't work:
driver.find_element_by_css_selector(".x1p[onclick*='details|1']").click()
Another option would be selecting the preceding element, with the following code:
precedingbutton = driver.find_element_by_name('B22_1_6')
And then sending tab and then enter. However, after I send Tab, I don't know of a way to send Enter without assigning the Send_Keys command back to the preceding box, which deselects the button I want.
Let me know if you can help!

If you can locate the parent, you can locate the child, either with xpath or the nth-child css selector, relative to it.
Edit in response to comments:
Assuming by "2 elements away" you mean it is a sibling preceding the target, try something like td[name="B22_1_6"] + td + td. This is the sibling selector.

You can do it easily by
driver.find_element(By.XPATH, 'your xpath').click()
For finding the xpath , just inspect the button with firebug, and in html tab right click on selected element and click on Copy XPath.
Hope it will help you.