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Selenium: Unable to locate element by class and id

Trying to scrape a website, I created a loop and was able to locate all the elements. My problem is, that the next button id changes on every page. So I can not use the id as a locator.
This is the next button on page 1:
<a rel="nofollow" id="f_c7" href="#" class="nextLink jasty-link"></a>
And this is the next button on page 2:
<a rel="nofollow" id="f_c9" href="#" class="nextLink jasty-link"></a>
Idea:
next_button = browser.find_elements_by_class_name("nextLink jasty-link")
next_button.click
I get this error message:
Message: no such element: Unable to locate element
The problem here might be that there are two next buttons on the page.
So I tried to create a list but the list is empty.
next_buttons = browser.find_elements_by_class_name("nextLink jasty-link")
print(next_buttons)
Any idea on how to solve my problem? Would really appreciate it.
This is the website:
https://fazarchiv.faz.net/faz-portal/faz-archiv?q=Kryptow%C3%A4hrungen&source=&max=10&sort=&offset=0&_ts=1657629187558#hitlist

There are two issues in my opinion:
Depending from where you try to access the site there is a cookie banner that will get the click, so you may have to accept it first:
browser.find_element_by_class_name('cb-enable').click()
To locate a single element, one of the both next buttons, it doeas not matter, use browser.find_element() instead of browser.find_elements().
Selecting your element by multiple class names use xpath:
next_button = browser.find_element(By.XPATH, '//a[contains(#class, "nextLink jasty-link")]')
or css selectors:
next_button = browser.find_element(By.CSS_SELECTOR, '.nextLink.jasty-link')
Note: To avoid DeprecationWarning: find_element_by_* commands are deprecated. Please use find_element() import in addition from selenium.webdriver.common.by import By

You can't get elements by multiple class names. So, you can use find_elements_by_css_selector instead.
next_buttons = browser.find_elements_by_css_selector(".nextLink.jasty-link")
print(next_buttons)
You can then loop through the list and click the buttons:
next_buttons = browser.find_elements_by_css_selector(".nextLink.jasty-link")
for button in next_buttons:
button.click()

Try below xPath
//a[contains(#class, 'step jasty-link')]/following-sibling::a

Get text from div using Selenium and Python

Situation
I'm using Selenium and Python to extract info from a page
Here is the div I want to extract from:
I want to extract the "Registre-se" and the "Login" text.
My code
from selenium import webdriver
url = 'https://www.bet365.com/#/AVR/B146/R^1'
driver = webdriver.Chrome()
driver.get(url.format(q=''))
elements = driver.find_elements_by_class_name('hm-MainHeaderRHSLoggedOutNarrow_Join ')
for e in elements:
print(e.text)
elements = driver.find_elements_by_class_name('hm-MainHeaderRHSLoggedOutNarrow_Login ')
for e in elements:
print(e.text)
Problem
My code don't send any output.
HTML
<div class="hm-MainHeaderRHSLoggedOutNarrow_Join ">Registre-se</div>
<div class="hm-MainHeaderRHSLoggedOutNarrow_Login " style="">Login</div>

By looking this HTML
<div class="hm-MainHeaderRHSLoggedOutNarrow_Join ">Registre-se</div>
<div class="hm-MainHeaderRHSLoggedOutNarrow_Login " style="">Login</div>
and your code, which looks okay to me, except that part you are using find_elements for a single web element.
and by reading this comment
The class name "hm-MainHeaderRHSLoggedOutMed_Login " only appear in
the inspect of the website, but not in the page source. What it's
supposed to do now?
It is clear that the element is in either iframe or shadow root.
Cause page_source does not look for iframe.
Please check if it is in iframe, then you'd have to switch to iframe first and then you can use the code that you have.
switch it like this :
driver.switch_to.frame(driver.find_element_by_xpath('xpath here'))

Problem in scraping link(href) using selenium; href="#"

I'm a amateur at using python, and I'm trying to scrape the url from the html below using selenium.
<a class="" href="#" style="text-decoration: none; color: #1b1b1b;" onclick="toDetailOrUrl(event, '1641438','')">[안내] 빗썸 - 빗썸 글로벌 간 간편 가상자산 이동 서비스 종료 안내</a>
In ordinary case, the link url i want to get is in just beside 'href=', but there is just "#" in that html.
When i run the code below that is usual way to using selenium to scrape the given html, it returns a https://cafe.bithumb.com/view/boards/43. But is just what i entered in 'driver.get()', and i don't want.
url = "https://cafe.bithumb.com/view/boards/43"
driver=webdriver.Chrome('chromedriver.exe')
driver.get(url)
driver.implicitly_wait(30)
bo =driver.find_element_by_xpath("//tbody[1]/tr[#style='cursor:pointer;border-top:1px solid #dee2e6;background-color: white']/td[2]/a")
print(bo.get_attribute('href'))
What i want is https://cafe.bithumb.com/view/board-contents/1641438. You can get this url when you click a item corresponding with the xpath i wrote above.
I want this url using selenium or other programmatic ways, no need to open a chrome and enter the url in addressbar, and click using mouse... like that.
good

You can use,
bo.click()
in order to click the element you want (I assumed you want to click bo)

print(driver.execute_script('return arguments[0].getAttribute("href")',bo))
selenium , bo.get_attribute('href') is actually doing document.getElementById("somelocaator").href which returns full href , as '#' indicates current page you get current URL you provided in get()
If you just need # you can use the execute_script

Selenium find_element_with_link_text not working

I have a piece of code as follows
<a class="country" href="/es-hn">
Honduras
</a>
and I'm trying to assign to to a variable by doing
el = self.driver.find_element_by_link_text('Honduras')
However whenever I run it I get the following error:
NoSuchElementException: Message: u"Unable to find element with link text == Honduras"

I've seen link_text fuzz up when trying to find a link when it's in block formation like this. I think it has something to do with the tabulation:
<a class="country" href="/es-hn">
[ ]Honduras
</a>
It only seems to work consistently when in line like this:
<a class="country" href="/es-hn">Honduras</a>
Try this:
el = self.driver.find_element_by_css_selector("a.country[href$='es-hn']")

I agree with sircapslot, in this case partial link text would also work:
el = self.driver.find_element_by_partial_link_text('Honduras')

This may occur when selenium is trying to find the link while your application hasn't rendered it yet. So you need to make it wait until the link appears:
browser.implicitly_wait(10) # 10 seconds
el = self.driver.find_element_by_link_text('Honduras')
The implicitly_wait call makes the browser poll until the item is on the page and visible to be interacted with.

Selenium in Python: Select the second element with given link text

I am trying to click all of the links on a web page that contain the link text "View all hits in this text." Here's what some of the html on the web page looks like:
<a href="/searchCom.do?offset=24981670&entry=4&entries=112&area=Poetry&forward=textsCom&queryId=../session/1380145118_2069"><b>View all hits in this text</b>
<br>
</a>
[...]
<a href="/searchCom.do?offset=25280103&entry=5&entries=112&area=Poetry&forward=textsCom&queryId=../session/1380145118_2069"><b>View all hits in this text</b>
<br>
</a>
If there were only one such link on the page, I know I could click it using something like:
driver.find_element_by_link_text('View all hits in this text').click()
Unfortunately, this method only ever identifies and clicks the first link on the web page with the link text "View all hits in this text." I therefore wanted to ask: is there a method I can use to click the second (or nth) link with link text "View all hits in this text" on this page? I have a feeling I may need to use xpath, but I haven't quite figured out how I should go about implementing xpath in my script. I would be grateful for any advice others can lend.

There is find_elements_by_link_text() (docs):
links = driver.find_elements_by_link_text('View all hits in this text')
for link in links:
link.click()
Also, you can use xpath to get all links with a specified text:
links = driver.find_elements_by_xpath("//a[text() = 'View all hits in this text']")
for link in links:
link.click()
Hope that helps.

Try this below code:
driver.find_elements_by_link_text('linktext')[1].click()