I am trying to crawl wordreference, but I am not succeding.
The first problem I have encountered is, that a big part is loaded via JavaScript, but that shouldn't be much problem because I can see what I need in the source code.
So, for example, I want to extract for a given word, the first two meanings, so in this url: http://www.wordreference.com/es/translation.asp?tranword=crane I need to extract grulla and grúa.
This is my code:
import lxml.html as lh
import urllib2
url = 'http://www.wordreference.com/es/translation.asp?tranword=crane'
doc = lh.parse((urllib2.urlopen(url)))
trans = doc.xpath('//td[#class="ToWrd"]/text()')
for i in trans:
print i
The result is that I get an empty list.
I have tried to crawl it with scrapy too, no success. I am not sure what is going on, the only way I have been able to crawl it is using curl, but that is sloopy, I want to do it in an elegant way, with Python.
Thank you very much
It looks like you need a User-Agent header to be sent, see Changing user agent on urllib2.urlopen.
Also, just switching to requests would do the trick (it automatically sends the python-requests/version User Agent by default):
import lxml.html as lh
import requests
url = 'http://www.wordreference.com/es/translation.asp?tranword=crane'
response = requests.get("http://www.wordreference.com/es/translation.asp?tranword=crane")
doc = lh.fromstring(response.content)
trans = doc.xpath('//td[#class="ToWrd"]/text()')
for i in trans:
print(i)
Prints:
grulla
grúa
plataforma
...
grulla blanca
grulla trompetera
Related
I'm new to Python, just get started with it today.
My system environment are Python 3.5 with some libraries on Windows10.
I want to extract football player data from site below as CSV file.
Problem: I can not extract data from soup.find_all('script')[17] to my expected CSV format. How to extract those data as I want ?
My code is shown as below.
from bs4 import BeautifulSoup
import re
from urllib.request import Request, urlopen
req = Request('http://www.futhead.com/squad-building-challenges/squads/343', headers={'User-Agent': 'Mozilla/5.0'})
webpage = urlopen(req).read()
soup = BeautifulSoup(webpage,'html.parser') #not sure if i need to use lxml
soup.find_all('script')[17] #My target data is in 17th
My expected output would be similar to this
position,slot_position,slug
ST,ST,paulo-henrique
LM,LM,mugdat-celik
As #josiah Swain said, it's not going to be pretty. For this sort of thing it's more recommended to use JS as it can understand what you have.
Saying that, python is awesome and here is you solution!
#Same imports as before
from bs4 import BeautifulSoup
import re
from urllib.request import Request, urlopen
#And one more
import json
# The code you had
req = Request('http://www.futhead.com/squad-building-challenges/squads/343',
headers={'User-Agent': 'Mozilla/5.0'})
webpage = urlopen(req).read()
soup = BeautifulSoup(webpage,'html.parser')
# Store the script
script = soup.find_all('script')[17]
# Extract the oneline that stores all that JSON
uncleanJson = [line for line in script.text.split('\n')
if line.lstrip().startswith('squad.register_players($.parseJSON') ][0]
# The easiest way to strip away all that yucky JS to get to the JSON
cleanJSON = uncleanJson.lstrip() \
.replace('squad.register_players($.parseJSON(\'', '') \
.replace('\'));','')
# Extract out that useful info
data = [ [p['position'],p['data']['slot_position'],p['data']['slug']]
for p in json.loads(cleanJSON)
if p['player'] is not None]
print('position,slot_position,slug')
for line in data:
print(','.join(line))
The result I get for copying and pasting this into python is:
position,slot_position,slug
ST,ST,paulo-henrique
LM,LM,mugdat-celik
CAM,CAM,soner-aydogdu
RM,RM,petar-grbic
GK,GK,fatih-ozturk
CDM,CDM,eray-ataseven
LB,LB,kadir-keles
CB,CB,caner-osmanpasa
CB,CB,mustafa-yumlu
RM,RM,ioan-adrian-hora
GK,GK,bora-kork
Edit: On reflection this is not the easiest code to read for a beginner. Here is a easier to read version
# ... All that previous code
script = soup.find_all('script')[17]
allScriptLines = script.text.split('\n')
uncleanJson = None
for line in allScriptLines:
# Remove left whitespace (makes it easier to parse)
cleaner_line = line.lstrip()
if cleaner_line.startswith('squad.register_players($.parseJSON'):
uncleanJson = cleaner_line
cleanJSON = uncleanJson.replace('squad.register_players($.parseJSON(\'', '').replace('\'));','')
print('position,slot_position,slug')
for player in json.loads(cleanJSON):
if player['player'] is not None:
print(player['position'],player['data']['slot_position'],player['data']['slug'])
So my understanding is that beautifulsoup is better for HTML parsing, but you are trying to parse javascript nested in the HTML.
So you have two options
Simply create a function that takes the result of soup.find_all('script')[17], loop and search the string manually for the data and extract it. You can even use ast.literal_eval(string_thats_really_a_dictionary) to make it even easier. This is may not be the best a approach but if you are new to python you might want to do it this just for practice.
Use the json library like in this example. or alternatively like this way. This is probably the better way to do.
forgive me, if if come straight out with it but python drives me nuts at something what seemed to be quite simple.
In a nutshell
I'm writing an extension for a musicvideo scraper which is responsible for getting the fanart backdrop.
Here is the URL:
github.com/MViDLibraryToolKit/.../APICaller
So I was able to call the Fanart.tv API and receiving the right json response. My problem is that i'm to dumb to collect the URLs under the Element "artistbackground"
I search the internet and found a very similar post here at stackoverflow but unluckily this was concerning python2,API V2 and a different category at fanart.tv so I was not able to take use out of it. Here it was
Anyway, here is my poor Try to collect URLs to list
# --------------------- Response Verarbeitung
# Ausgabe zwecks Debug
# print(fanartTVresp)
# http://webservice.fanart.tv/v3/music/albums/ba853904-ae25-4ebb-89d6-c44cfbd71bd2?api_key=fdadba00cfaaf3621eaa748669256a9e&client_key=dce01d75553d7e3fbc2ad742aaf5d371
# zu befüllende Liste
url_list = []
# lade Web-Response
json_response = json.loads(fanartTVresp)
# durch Element artistbackground loopen
for artistbackground in json_response:
url = urllib.parse.quote(['url'], ':/')
if url:
url_list.append(url)
print(url_list)
The libs I loaded...
import musicbrainzngs
import urllib
import json
import socket
from pprint import pprint
from urllib.parse import quote
The rest from the code you can find at my github link. Please help me, it drives me crazy ^^
Kind regards
p.s. Please excuse my english, I came from germany :)
I think I finally got it.
# URL List for background images
url_list = []
# set only for debug / value came from powershell runtime later
location = os.path.abspath('C:/temp')
# decode json
json_response = json.loads(fanartTVresp.decode())
# set string objects
bgitem = json_response["artistbackground"]
bgcoverurl = json_response["artistbackground"][0]["url"]
# iterating items and collect
for bgcoverurl in bgitem:
url_list.append(bgcoverurl)
print(url_list)
After taking some hourse of sleep I reallized that "json.loads" deserialized the response to regular python objects. Correct me if I'm wrong.
Anyway, it finally works!
I'm currently working on an arcbot and I'm trying to make a command "!urbandictionary", it should scrape the meaning of a term, the first one which is provided by urbandictionary, if there's another solution, e.g. another dictionary site with a better api that's also good. Here's my code:
if Command.lower() == '!urban':
dictionary = Argument[1] #this is the term which the user provides, e.g. "scrape"
dictionaryscrape = urllib2.urlopen('http://www.urbandictionary.com/define.php?term='+dictionary).read() #plain html of the site
scraped = getBetweenHTML(dictionaryscrape, '<div class="meaning">','</div>') #Here's my problem, i'm not sure if it scrapes the first meaning or not..
messages.main(scraped,xSock,BotID) #Sends the meaning of the provided word (Argument[0])
How do I correctly scrape a meaning of a word in urbandictionary?
Just get the text from the meaning class:
import requests
from bs4 import BeautifulSoup
word = "scrape"
r = requests.get("http://www.urbandictionary.com/define.php?term={}".format(word))
soup = BeautifulSoup(r.content)
print(soup.find("div",attrs={"class":"meaning"}).text)
Gassing and breaking your car repeatedly really fast so that the front and rear bumpers "scrape" the pavement; while going hyphy
There is an unofficial api here apparently
`http://api.urbandictionary.com/v0/define?term={word}`
From https://github.com/zdict/zdict/wiki/Urban-dictionary-API-documentation
So I have a data retrieval/entry project and I want to extract a certain part of a webpage and store it in a text file. I have a text file of urls and the program is supposed to extract the same part of the page for each url.
Specifically, the program copies the legal statute following "Legal Authority:" on pages such as this. As you can see, there is only one statute listed. However, some of the urls also look like this, meaning that there are multiple separated statutes.
My code works for pages of the first kind:
from sys import argv
from urllib2 import urlopen
script, urlfile, legalfile = argv
input = open(urlfile, "r")
output = open(legalfile, "w")
def get_legal(page):
# this is where Legal Authority: starts in the code
start_link = page.find('Legal Authority:')
start_legal = page.find('">', start_link+1)
end_link = page.find('<', start_legal+1)
legal = page[start_legal+2: end_link]
return legal
for line in input:
pg = urlopen(line).read()
statute = get_legal(pg)
output.write(get_legal(pg))
Giving me the desired statute name in the "legalfile" output .txt. However, it cannot copy multiple statute names. I've tried something like this:
def get_legal(page):
# this is where Legal Authority: starts in the code
end_link = ""
legal = ""
start_link = page.find('Legal Authority:')
while (end_link != '</a> '):
start_legal = page.find('">', start_link+1)
end_link = page.find('<', start_legal+1)
end2 = page.find('</a> ', end_link+1)
legal += page[start_legal+2: end_link]
if
break
return legal
Since every list of statutes ends with '</a> ' (inspect the source of either of the two links) I thought I could use that fact (having it as the end of the index) to loop through and collect all the statutes in one string. Any ideas?
I would suggest using BeautifulSoup to parse and search your html. This will be much easier than doing basic string searches.
Here's a sample that pulls all the <a> tags found within the <td> tag that contains the <b>Legal Authority:</b> tag. (Note that I'm using requests library to fetch page content here - this is just a recommended and very easy to use alternative to urlopen.)
import requests
from BeautifulSoup import BeautifulSoup
# fetch the content of the page with requests library
url = "http://www.reginfo.gov/public/do/eAgendaViewRule?pubId=200210&RIN=1205-AB16"
response = requests.get(url)
# parse the html
html = BeautifulSoup(response.content)
# find all the <a> tags
a_tags = html.findAll('a', attrs={'class': 'pageSubNavTxt'})
def fetch_parent_tag(tags):
# fetch the parent <td> tag of the first <a> tag
# whose "previous sibling" is the <b>Legal Authority:</b> tag.
for tag in tags:
sibling = tag.findPreviousSibling()
if not sibling:
continue
if sibling.getText() == 'Legal Authority:':
return tag.findParent()
# now, just find all the child <a> tags of the parent.
# i.e. finding the parent of one child, find all the children
parent_tag = fetch_parent_tag(a_tags)
tags_you_want = parent_tag.findAll('a')
for tag in tags_you_want:
print 'statute: ' + tag.getText()
If this isn't exactly what you needed to do, BeautifulSoup is still the tool you likely want to use for sifting through html.
They provide XML data over there, see my comment. If you think you can't download that many files (or the other end could dislike so many HTTP GET requests), I'd recommend asking their admins if they would kindly provide you with a different way of accessing the data.
I have done so twice in the past (with scientific databases). In one instance the sheer size of the dataset prohibited a download; they ran a SQL query of mine and e-mailed the results (but had previously offered to mail a DVD or hard disk). In another case, I could have done some million HTTP requests to a webservice (and they were ok) each fetching about 1k bytes. This would have taken long, and would have been quite inconvenient (requiring some error-handling, since some of these requests would always time out) (and non-atomic due to paging). I was mailed a DVD.
I'd imagine that the Office of Management and Budget could possibly be similar accomodating.
Looking for a python script that would simply connect to a web page (maybe some querystring parameters).
I am going to run this script as a batch job in unix.
urllib2 will do what you want and it's pretty simple to use.
import urllib
import urllib2
params = {'param1': 'value1'}
req = urllib2.Request("http://someurl", urllib.urlencode(params))
res = urllib2.urlopen(req)
data = res.read()
It's also nice because it's easy to modify the above code to do all sorts of other things like POST requests, Basic Authentication, etc.
Try this:
aResp = urllib2.urlopen("http://google.com/");
print aResp.read();
If you need your script to actually function as a user of the site (clicking links, etc.) then you're probably looking for the python mechanize library.
Python Mechanize
A simple wget called from a shell script might suffice.
in python 2.7:
import urllib2
params = "key=val&key2=val2" #make sure that it's in GET request format
url = "http://www.example.com"
html = urllib2.urlopen(url+"?"+params).read()
print html
more info at https://docs.python.org/2.7/library/urllib2.html
in python 3.6:
from urllib.request import urlopen
params = "key=val&key2=val2" #make sure that it's in GET request format
url = "http://www.example.com"
html = urlopen(url+"?"+params).read()
print(html)
more info at https://docs.python.org/3.6/library/urllib.request.html
to encode params into GET format:
def myEncode(dictionary):
result = ""
for k in dictionary: #k is the key
result += k+"="+dictionary[k]+"&"
return result[:-1] #all but that last `&`
I'm pretty sure this should work in either python2 or python3...
What are you trying to do? If you're just trying to fetch a web page, cURL is a pre-existing (and very common) tool that does exactly that.
Basic usage is very simple:
curl www.example.com
You might want to simply use httplib from the standard library.
myConnection = httplib.HTTPConnection('http://www.example.com')
you can find the official reference here: http://docs.python.org/library/httplib.html