Python 3.5 | 从网站抓取数据

Question

I want to scrape a specific part of the website Kickstarter.com

I need the strings of the Project-title. The website is structured and every project has this line.

 <div class="Project-title"> 

My code looks like:

 #Loading Libraries import urllib import urllib.request from bs4 import BeautifulSoup #define URL for scraping theurl = "https://www.kickstarter.com/discover/advanced?category_id=16&woe_id=23424829&sort=popularity&seed=2448324&page=1" thepage = urllib.request.urlopen(theurl) #Cooking the Soup soup = BeautifulSoup(thepage,"html.parser") #Scraping "Project Title" (project-title) project_title = soup.find('h6', {'class': 'project-title'}).findChildren('a') title = project_title[0].text print (title) 

If I use the soup.find_all or set another value at the line Project_title[0] instead of zero, Python shows an error.

I need a list with all the project titles of this Website. Eg.:

• The Superbook: Turn your smartphone into a laptop for $99
• Weights: Weigh Smarter
• Mine Kafon Drone World's First And Only Complete
• Weather Camera System Omega2: $5 IoT Computer with Wi-Fi, Powered by Linux

Answer 1

find() only returns one element. To get all, you must use findAll

Here's the code you need

project_elements = soup.findAll('h6', {'class': 'project-title'})
project_titles = [project.findChildren('a')[0].text for project in project_elements]
print(project_titles)

We look at all the elements of tag h6 and class project-title . We then take the title from each of these elements, and create a list with it.

Hope it helped, and don't hesitate to ask if you have any question.

edit : the problem of the above code is that it will fail if we do not get at least a child of tag a for each element in the list returned by findAll

How to prevent this :

project_titles = [project.findChildren('a')[0].text for project in project_elements if project.findChildren('a')]

this will create the list only if the project.findChildren('a') as at least one element. ( if [] returns False)

edit : to get the description of the elements (class project-blurb ), let's look a bit at the HTML code.

<p class="project-blurb">
Bagel is a digital tape measure that helps you measure, organize, and analyze any size measurements in a smart way.
</p>

This is only a paragraph of class project-blurb . To get them, we could use the same as we did to get the project_elements, or more condensed :

project_desc = [description.text for description in soup.findAll('p', {'class': 'project-blurb'})]

Answer 2

With respect to the title of this post i would recommend you two different tutorial based on scraping particular data from a website . They do have a detailed explanation regarding how the task is achieved.

Firstly i would recommend to checkout pyimagesearch Scraping images using scrapy.

then you should try if you are more specific web scraping will help you.

Answer 3

All the data you want is in the section with the css class staff-picks , just find the h6's with the project-title class and extract the text from the anchor tag inside:

soup = BeautifulSoup(thepage,"html.parser")


print [a.text for a in soup.select("section.staff-picks h6.project-title a")]

Output:

[u'The Superbook: Turn your smartphone into a laptop for $99', u'Weighitz: Weigh Smarter', u'Omega2: $5 IoT Computer with Wi-Fi, Powered by Linux', u"Bagel: The World's Smartest Tape Measure", u'FireFlies - Truly Wire-Free Earbuds - Music Without Limits!', u'ISOLATE\xae - Switch off your ears!']

Or using find with find_all :

project_titles = soup.find("section",class_="staff-picks").find_all("h6", "project-title")
print([proj.a.text for proj in project_titles])

There is also only one anchor tag inside each h6 tag so you cannot end up with more than one whatever approach you take.