Develop Reference

Scrapy css selector get text from occurence of first class

I'm trying to scrape text from class .s-recipe-header__info-item, but as you can see on the picture, there are three classes with the same name and I would like to extract only the first one to get text "Do hodiny" See the image of code here. So far I have tried this code:
recipe_item["preparation_time"] = response.css(".s-recipe-header__info > .s-recipe-header__info-items > .s-recipe-header__info-item::text").extract_first()
I have also tried to use .get() instead of .extract_first(), but both do not seem to work...
I am new to web scraping and I have only elemental HTML and CSS knowledge. Thank you in advance for your help.

Python selenium crawling

Here is code
driver = webdriver.Chrome()
driver.get('https://tieba.baidu.com/f?kw=比特币&ie=utf-8&tab=good')
driver.find_elements_by_css_selector('a.j_th_tit')[0].click()
a = driver.find_elements_by_css_selector('div.d_post_content.j_d_post_content.clearfix')
for i in a:
print(i.text)
Here is HTML I'm struggling with. There are many texts at the page, but those all have same class; d_post_content j_d_post_content clearfix.
<div id='post_content_52497574149' class='d_post_content j_d_post_content clearfix' style='display:;'> Here is the Text that I need to get; it is written in Chinese and stackoverflow may not permit to writhe Chinese in the body </div>
I want to automatically access to the website and get some texts for my homework assignment. With this code above, I could open the website, click the link, but I cannot access to the text needed. All of the texts needed are in the class, so I tried to access to the Class to get the texts, but it didn't work. When I check the length of the list a, len(a) is zero. Could anyone help me?

This line bring you to a new tab:
driver.find_elements_by_css_selector('a.j_th_tit')[0].click()
So you need switch it first. After perform the above, just add a line:
driver.switch_to.window(driver.window_handles[-1])

When you click the link in this statement:
driver.find_elements_by_css_selector('a.j_th_tit')[0].click()
A new tab is opened. But you are not switching to that tab.
I would recommend adding this statement:
driver.switch_to.window(driver.window_handles[-1])
Before you actually call find_elements_by_css_selector.
It will solve your issue.

remove css style code when I get a webpage text

I would like to get full text of a webpage, unfortunately my scraper is also capturing css code, how i can complete the code below in order to remove also css style code :
page = " ".join(response.xpath('//body//descendant-or-self::*[not(self::script)]/text()').extract())

Try
//body//descendant-or-self::*[not(self::script or self::style)]
I tested and it works, it excludes STYLE and SCRIPT tags

How to use Selenium to input text in Python

I want to input "apple" instead of "hello" using Selenium in Python.
<span><div class="top1"><div class="top2"><label class="top3">
<label class="check"><input class="word" type="text" value="hello"></label>
How to do it?
Thank you very much.

As per the HTML you have provided, to send the text apple within the <input> element you can use the following code block :
driver.find_element_by_xpath("//label[#class='check']/input[#class='word' and #type='text' and #value='hello']").click()
driver.find_element_by_xpath("//label[#class='check']/input[#class='word' and #type='text' and #value='hello']").clear()
driver.find_element_by_xpath("//label[#class='check']/input[#class='word' and #type='text' and #value='hello']").send_keys("apple")

Before asking elementary questions like this, please read documentation for tools which you are using. You can find documentation via Google.
About your task. First, you need find your input in source code, than clean value and input a new one.
your_input = driver.find_element_by_xpath('xpath to your input')
your_input.clear()
your_input.send_keys('your new text')

Using Python and Selenium why am I unable to find link by link text?

I have a list webelement that has a bunch of links within it. The html looks like:
<li>
<span class="ss-icon"></span> Remove
<a href="/sessions/new"><span class="ss-icon"></span> Sign in to save items</a
...
When I try to do something like:
link = element.find_element_by_link_text('Sign in to save items')
I get an error that says:
NoSuchElementException: Message: Unable to locate element:
{"method":"link text","selector":"Sign in to save items"}
I have been able to find this link by instead doing a find_elements_by_tag_name('a') and then just using the link with the correct HREF, but I would like to understand why the first method fails.

It happened to me before that the find_element_by_link_text method sometimes works and sometimes doesn't work; even in a single case. I think it's not a reliable way to access elements; the best way is to use find_element_by_id.
But in your case, as I visit the page, there is no id to help you. Still you can try find_elements_by_xpath in 2 ways:
1- Accessing title: find_element_by_xpath["//a[contains(#title = 'Sign in to save items')]"]
2- Accessing text: find_element_by_xpath["//a[contains(text(), 'Sign in to save items')]"]
Hope it helps.

The problem is, most likely, in the extra spaces before or/and after the link text. You can still approach it with a "partial link text' match:
element.find_element_by_partial_link_text('Sign in to save items')