I am VERY new to web scraping in any shape or form, I've been trying to get into Python and I heard that web scraping was a good way to expose myself to Python. So, after many Google searches I finally came down to the use of two highly recommended modules: Requests and BeautifulSoup. I've read up a fair amount on both and have a basic understanding on how to use them.
I found a very basic website (basic in that there isn't much content or javascript and the like, making parsing the HTML a lot easier) and I have the following code:
import requests
from bs4 import BeautifulSoup
soup = BeautifulSoup(requests.get('http://www.basicwebs.co.uk/contact.htm').text)
for row in soup('div',{'id': 'Layer1'})[0].h2('font'):
tds = row.text
print tds
This code works. It produces the following result:
BASIC
WEBS
Contact details
Contact details
Which, if you spend a few minutes inspecting the code on this page, is the correct result (I assume). Now, the thing is, while this code works, what if I wanted to get a different part of the page? Like the little paragraph on the page that states "If you are interested in having a website designed and hosted by us, please contact us either by e-mail or telephone." - my understanding would be to simply change the index number to the corresponding header that this text is found under, but when I change it I get a message that the list index is out of range.
Can anybody help? (as simple as you can make it, if possible)
I'm using Python 2.7.8
The text you require surrounded by the font tag with an attribute size=3, so one way to do it is by selecting the first occurrence of it like this:
font_elements = soup('font', {'size': 3})
if font_elements:
print font_elements[0].text
RESULT:
If you are interested in having a website designed
and hosted by us, please contact us either by e-mail or telephone.
You can directly do this :
soup('font',{'size': '3'})[0].text
However, I want to draw your attention towards the mistake you made before.
soup('div',{'id': 'Layer1'})
this returns the div tag with id='Layer1' which can be more than one. So it basically returns a list of all HTML elements whose div tags have id='Layer1' but unfortunately the HTML you were trying to parse has one such element. So it went out of bound.
You can probably use some interactive interpreter of python like bpython or ipython to test what are you getting in an object.? Happy Hacking!!!
from urllib.request import urlopen
from bs4 import BeautifulSoup
web_address=' http://www.basicwebs.co.uk/contact.htm'
html = urlopen(web_address)
bs = BeautifulSoup(html.read(), 'html.parser')
contact_info = bs.findAll('h2', {'align':'left'})[0]
for info in contact_info:
print(info.get_text())
Related
I try to do some Web Scraping
The objective is to collect all remedials according to the postal code. The problem is when I try my code, my list is empty because the url did't change according to the postal code. This is why I want to change the HTML value during the scrape.
I'm not sure how to do this. I tried using Selenium and XPATH however I wasn't able to find anything.
Here's the HTML Code: (in red is what I need to change.)
EDIT : Indeed, the goal is to collect the pagination with the name and the type of remedial according to the postal code, this is why I want to change the HTML content during the scrap.
This is the best that I can do for the moment, I hope u will see the error
This input is in a form, which is good because Selenium has special functionalities to handle forms.
from selenium import webdriver
url = "https://www.maif.fr/services-en-ligne/consultationreparateurs/geolocaliserReparateur.action?view"
query = "whatever you want to put into the search box"
driver = webdriver.Chrome()
driver.get(url)
webform_input = driver.find_element_by_xpath("//input[#id='adresseInternaute']")
webform_input.send_keys(query)
webform_input.submit()
The key here is submit(). It will walk the HTML tree until it finds a button within the current form, meaning you don't have to write an extra two lines just to click the search button.
I need to scrape this page to get the value of the comment, as well as the Document and Submitter Information on the right side..
https://www.regulations.gov/document?D=FDA-2014-N-1207-7673
I've tried using read_html() and read_xml() from the xml2 package with no luck. I've tried getURLContent() followed by xmlParse() and htmlParse() from RCurl.
I even tried simply readLines(), which does not actually get me the content of the website.
I suppose I don't have a great understanding how this all works. Previous websites I have always been able to scrape with simply html_parse(), html_nodes() and html_attr(). How can I accomplish scraping this website?
I am working on a project where I want to scrape a page like this, in order to get the city of origin. I tried to use the css selector: ".type-12~ .type-12+ .type-12" However I do not get the text into R.
Link:
https://www.kickstarter.com/projects/1141096871/support-ctrl-shft/description
I use rvest and and the read_html function.
However, it seems that the source has some scripts in it. Is there a way to scrape the website after the scripts have returned their results (as you see it with a browser)?
PS I looked at similar questions but did find the answer..
Code:
main.names <- read_html(x = paste0("https://www.kickstarter.com/projects/1141096871/support-ctrl-shft/description")) # feed `main.page` to the next step
names1 <- main.names %>% # feed `main.page` to the next step
html_nodes("div.mb0-md") %>% # get the CSS nodes
html_text()# extract the text
You should not do it. They provide a API which you can find here: https://status.kickstarter.com/api
Using APIs or Ajax/JSON calls is usually better since
The server isn't overused because your scraper visits every link it can find causing unnecessary traffic. That is bad for the speed of your program and bad for the servers of the site you are scraping.
You don't have to worry about that they changed a class name or id and your code won't work anymore
Especially the second part should interest you since it can take hours finding which class isn't returning a value anymore.
But to answer your question:
When you use the right scraper you can find all what you want. What tools are you using? There are possibilities to get data before the site is loaded or after. You can execute the JS on the site separately and find hidden content or find things like display:none Css classes...
It really depends on what you are using and how you use it.
I'm not great in HTML, so am a bit stumbled for this.
I'm trying to scrape instagram datetime posts using python, and realised that the datetime information isn't without the html document of the post. However, I am able to query it using inspect element. See below screen shot.
Where is this datetime information located exactly, and how can I obtain it?
The example I took from is this random post "https://www.instagram.com/p/BEtMWWbjoPh/". Element is at the "12h" displayed in the page.
[Update] I am using urllib to grab the url, and bs4 in python to scrape. The output did not return anything with datetime. The code is below. I also printed out the entire html and I was surprised that it does not contain datetime in it.
html = urllib.urlopen(url).read()
soup = BeautifulSoup(html, 'html.parser')
tags = soup.select('time')
for tag in tags:
dateT = tag.get('datetime').getText()
print dateT
In your developer console, type this:
document.getElementsByTagName('time')[0].getAttribute('datetime');
This will return the data you are looking for. The above code is simply looking through the HTML for the tag name time, of which there is only one, then grabbing the datetime property from it.
As for python, check out BeautifulSoup if you haven't already. This library will allow you to do a similar thing in python:
from bs4 import BeautifulSoup
soup = BeautifulSoup(html_doc, 'html.parser')
soup.time['datetime']
Where html_doc is your raw HTML. To obtain the raw HTML, use the requests library.
I think the problem that you are experiencing is that urllib.urlopen(url).read() does not execute any javascript that is on the page.
Because Instagram is a client side javascript app that uses your browser to render their site, you'll need some sort of browser client to evaluate the javascript and then find the element on the page. For this, I usually use phantomjs (I usually use it with the ruby driver Capybara, but I would assume that there is a python package that would work similarly)
HOWEVER, if you execute urllib.urlopen(url).read(), you should see a block of JSON in a script tag that begins with <script type="text/javascript">window._sharedData = {...
That block of JSON will include the data you are looking for. If you were to evaluate that JSON, and parse it, you should be able access the time data you are looking for.
That being said, the better way to do this is to use instagram's api to do the the crawling. They make all of this data available to developers, so you don't have to crawl an ever-changing webpage.
(Apparently Instagram's API will only return public data for users who have explicitly given your app permission)
hi so I need to retrieve the url for the first article on a term I search up on nytimes.com
So if I search for Apple. This link would return the result
http://query.nytimes.com/search/sitesearch?query=Apple&srchst=cse
And you just replace Apple with the term you are searching for.
If you click on that link you would see that NYtimes ask you if you mean Apple Inc.
I want to get the url for this link, and go to it.
Then you will just get a lot of information on Apple Inc.
If you scroll down you will see the articles related to Apple.
So what I ultimately want is the URL of the first article on this page.
So I really do not know how to go about this. Do I use Java, or what do I use? Any help would be greatly appreciated and I would put a bounty on this later, but I need the answer ASAP.
Thanks
EDIT: Can we do this in Java?
You can use Python with the standard urllib module to fetch the pages and the great HTML parser BeautifulSoup to obtain the information you need from the pages.
From the documentation of BeautifulSoup, here's sample code that fetches a web page and extracts some info from it:
import urllib2
from BeautifulSoup import BeautifulSoup
page = urllib2.urlopen("http://www.icc-ccs.org/prc/piracyreport.php")
soup = BeautifulSoup(page)
for incident in soup('td', width="90%"):
where, linebreak, what = incident.contents[:3]
print where.strip()
print what.strip()
print
This this is a nice and detailed article on the topic.
You certainly can do it in Java. Look at the HttpURLConnection class. Basically, you give it a URL, call the connect function, and you get back an input stream with the contents of the page, i.e. HTML text. You can then process that and parse out whatever information you want.
You're facing two challenges in the project you are describing. The first, and probably really the lesser challenge, is figuring out the mechanics of how to connect to a web page and get hold of the text within your program. The second and probably bigger challenge will be to figure out exactly how to extract the information you want from that text. I'm not clear on the details of your requirements, but you're going to have to sort through a ton of text to find what you're looking for. Without actually looking at the NY Times site at the momemnt, I'm sure it has all sorts of decorations like pretty pictures and the company logo and headlines and so on, and then there are going to be menus and advertisements and all sorts of stuff. I sincerely doubt that the NY Times or almost any other commercial web site is going to return a search page that includes nothing but a link to the article you are interested in. Somehow your program will have to figure out that the first link is to the "subscribe on line" page, the second is to an advertisement, the third is to customer service, the fourth and fifth are additional advertisements, the sixth is to the home page, etc etc until you finally get to the one you're actually interested in. How will you identify the interesting link? There are probably headings or formatting that make it recognizable to a human being, but you use a lot of intuition to screen out the clutter that can be difficult to reproduce in a program.
Good luck!
You can do this in C# using the HTML Agility Pack, or using LINQ to XML if the site is valid XHTML. EDIT: It isn't valid XHTML; I checked.
The following (tested) code will get the URL of the first search result:
var doc = new HtmlWeb().Load(#"http://query.nytimes.com/search/sitesearch?query=Apple&srchst=cse");
var url = HtmlEntity.DeEntitize(doc.DocumentNode.Descendants("ul")
.First(ul => ul.Attributes["class"] != null
&& ul.Attributes["class"].Value == "results")
.Descendants("a")
.First()
.Attributes["href"].Value);
Note that if their website changes, this code might stop working.