I am trying to save all the < a > links within the python homepage into a folder named 'Downloaded pages'. However after 2 iterations through the for loop I receive the following error:
www.python.org#content <_io.BufferedWriter name='Downloaded
Pages/www.python.org#content'> www.python.org#python-network
<_io.BufferedWriter name='Downloaded
Pages/www.python.org#python-network'>
Traceback (most recent call last): File "/Users/Lucas/Python/AP book
exercise/Web Scraping/linkVerification.py", line 26, in
downloadedPage = open(os.path.join('Downloaded Pages', os.path.basename(linkUrlToOpen)), 'wb') IsADirectoryError: [Errno 21]
Is a directory: 'Downloaded Pages/'
I am unsure why this happens as it appears the pages are being saved as due to seeing '<_io.BufferedWriter name='Downloaded Pages/www.python.org#content'>', which says to me its the correct path.
This is my code:
import requests, os, bs4
# Create a new folder to download webpages to
os.makedirs('Downloaded Pages', exist_ok=True)
# Download webpage
url = 'https://www.python.org/'
res = requests.get(url)
res.raise_for_status() # Check if the download was successful
soupObj = bs4.BeautifulSoup(res.text, 'html.parser') # Collects all text form the webpage
# Find all 'a' links on the webpage
linkElem = soupObj.select('a')
numOfLinks = len(linkElem)
for i in range(numOfLinks):
linkUrlToOpen = 'https://www.python.org' + linkElem[i].get('href')
print(os.path.basename(linkUrlToOpen))
# save each downloaded page to the 'Downloaded pages' folder
downloadedPage = open(os.path.join('Downloaded Pages', os.path.basename(linkUrlToOpen)), 'wb')
print(downloadedPage)
if linkElem == []:
print('Error, link does not work')
else:
for chunk in res.iter_content(100000):
downloadedPage.write(chunk)
downloadedPage.close()
Appreciate any advice, thanks.
The problem is that when you try to do things like parse the basename of a page with an .html dir it works, but when you try to do it with one that doesn't specify it on the url like "http://python.org/" the basename is actually empty (you can try printing first the url and then the basename bewteen brackets or something to see what i mean). So to work arround that, the easiest solution would be to use absolue paths like #Thyebri said.
And also, remember that the file you write cannot contain characters like '/', '\' or '?'
So, i dont know if the following code it's messy or not, but using the re library i would do the following:
filename = re.sub('[\/*:"?]+', '-', linkUrlToOpen.split("://")[1])
downloadedPage = open(os.path.join('Downloaded_Pages', filename), 'wb')
So, first i remove part i remove the "https://" part, and then with the regular expressions library i replace all the usual symbols that are present in url links with a dash '-' and that is the name that will be given to the file.
Hope it works!
Related
this is an URL example "https://procurement-notices.undp.org/view_file.cfm?doc_id=257280"
if you put it in the browser a file will start downloading in your system.
I want to download this file using python and store it somewhere on my computer
this is how tried
import requests
# first_url = 'https://readthedocs.org/projects/python-guide/downloads/pdf/latest/'
second_url="https://procurement-notices.undp.org/view_file.cfm?doc_id=257280"
myfile = requests.get(second_url , allow_redirects=True)
# this works for the first URL
# open('example.pdf' , 'wb').write(myfile.content)
# this did't work for both of them
# open('example.txt' , 'wb').write(myfile.content)
# this works for the second URL
open('example.doc' , 'wb').write(myfile.content)
first: if I put the first_url in the browser it will download a pdf file, putting second_url will download a .doc file How can I know what type of file will the URL give to us or what type of file will be downloaded so that I use the correct open(...) method?
second: If I use the second URL in the browser a file with the name "T__proc_notices_notices_080_k_notice_doc_79545_770020123.docx" starts downloading. how can I know this file name when I try to download the file?
if you know any better solution kindly let me know for the implementation.
kindly have a quick look at Downloading Files from URLs and zip downloaded files in python question aswell
myfile.headers['content-type'] will give you the MIME-type of the URL's content and myfile.headers['content-disposition'] gives you info like filename etc. (if the response contains this header at all)
you can use response headers content-type like for first url it is application/pdf and sencond url for is application/msword you save file according to it. you can make extension dictinary where you can store possible file format and their types and match with it. your second question is also same like this one so i am taking your two urls from that question and for file name i am using just integers
all_Urls = ['https://omextemplates.content.office.net/support/templates/en-us/tf16402488.dotx' ,
'https://procurement-notices.undp.org/view_file.cfm?doc_id=257280']
extension_dict = {'application/vnd.openxmlformats-officedocument.wordprocessingml.document':'.docx',
'application/vnd.openxmlformats-officedocument.wordprocessingml.template':'.dotx',
'application/vnd.ms-word.document.macroEnabled.12':'.docm',
'application/vnd.ms-word.template.macroEnabled.12':'.dotm',
'application/pdf':'.pdf',
'application/msword':'.doc'}
for i,url in enumerate(all_Urls):
resp = requests.get(url)
response_headers = resp.headers
file_extension = extensio_dict[response_headers['Content-Type']]
with open(f"{i}.{file_extension}",'wb') as f:
f.write(resp.content)
for MIME-Type see this answer
I am writing small python code to download a file from follow link and retrieve original filename
and its extension.But I have come across one such follow link for which python downloads the file but it is without any extension whereas file has .txt extension when downloads using browser.
Below is the code I am trying :
from urllib.request import urlopen
from urllib.parse import unquote
import wget
filePath = 'D:\\folder_path'
followLink = 'http://example.com/Reports/Download/c4feb46c-8758-4266-bec6-12358'
response = urlopen(followLink)
if response.code == 200:
print('Follow Link(response url) :' + response.url)
print('\n')
unquote_url = unquote(response.url)
file_name = wget.detect_filename(response.url).replace('|', '_')
print('file_name - '+file_name)
wget.download(response.url,filePa
th)
file_name variable in above code is just giving 'c4feb46c-8758-4266-bec6-12358' as filename.
Where I want to download it as c4feb46c-8758-4266-bec6-12358.txt.
I have also tried to read file name from header i.e. response.info(). But not getting proper file name.
Anyone can please help me with this.I am stucked in my work.Thanks in advance.
Wget gets the filename from the URL itself. For example, if your URL was https://someurl.com/filename.pdf, it is saved as filename.pdf. If it was https://someurl.com/filename, it is saved as filename. Since wget.download returns the filename of the downloaded file, you can rename it to any extension you want with os.rename(filename, filename+'.<extension>').
Ok so I am using a script that is downloading a files from urls listed in a urls.txt.
import urllib.request
with open("urls.txt", "r") as file:
linkList = file.readlines()
for link in linkList:
urllib.request.urlretrieve(link)
Unfortunately they are saved as temporary files due to lack of second argument in my urllib.request.urlretrieve function. As there are thousand of links in my text file naming them separately is not an option. The thing is that the name of the file is contained in those links, i.e. /DocumentXML2XLSDownload.vm?firsttime=true&repengback=true&documentId=XXXXXX&xslFileName=rher2xml.xsl&outputFileName=XXXX_2017_06_25_4.xls where the name of the file comes after outputFileName=
Is there an easy way to parse the file names and then use them in urllib.request.urlretrieve function as secondary argument? I was thinking of extracting those names in excel and placing them in another text file that would be read in similar fashion as urls.txt but I'm not sure how to implement it in Python. Or is there a way to make it exclusively in python without using excel?
You could parse the link on the go.
Example using a regular expression:
import re
with open("urls.txt", "r") as file:
linkList = file.readlines()
for link in linkList:
regexp = '((?<=\?outputFileName=)|(?<=\&outputFileName=))[^&]+'
match = re.search(regexp, link.rstrip())
if match is None:
# Make the user aware that something went wrong, e.g. raise exception
# and/or just print something
print("WARNING: Couldn't find file name in link [" + link + "]. Skipping...")
else:
file_name = match.group(0)
urllib.request.urlretrieve(link, file_name)
You can use urlparse and parse_qs to get the query string
from urlparse import urlparse,parse_qs
parse = urlparse('http://www.cwi.nl:80/%7Eguido/Python.html?name=Python&version=2')
print(parse_qs(parse.query)['name'][0]) # prints Python
I developed a web crawler to extract all the source codes in a wiki link. The program terminates after writing a few files.
def fetch_code(link_list):
for href in link_list:
response = urllib2.urlopen("https://www.wikipedia.org/"+href)
content = response.read()
page = open("%s.html" % href, 'w')
page.write(content.replace("[\/:?*<>|]", " "))
page.close()
link_list is an array, which has the extracted links from the seed page.
The error I get after executing is
IOError: [Errno 2] No such file or directory: u'M/s.html'
you cannot create a file with '/' in its name.
you could escape the filename as M%2Fs.html
/ is %2F
in python2, you could simply use urllib to escape the filename, example:
import urllib
filePath = urllib.quote_plus('M/s.html')
print(filePath)
on the other hand, you could also save http response to hierarchy, for example, M/s.html means s.html file under directory named 'M'.
I want to save all images from a site. wget is horrible, at least for http://www.leveldesigninspirationmachine.tumblr.com since in the image folder it just drops html files, and nothing as an extension.
I found a python script, the usage is like this:
[python] ImageDownloader.py URL MaxRecursionDepth DownloadLocationPath MinImageFileSize
Finally I got the script running after some BeautifulSoup problems.
However, I can't find the files anywhere. I also tried "/" as the output dir in hope the images got on the root of my HD but no luck. Can someone either help me to simplify the script so it outputs at the cd directory set in terminal. Or give me a command that should work. I have zero python experience and I don't really want to learn python for a 2 year old script that maybe doesn't even work the way I want.
Also, how can I pass an array of website? With a lot of scrapers it gives me the first few results of the page. Tumblr has the load on scroll but that has no effect so i would like to add /page1 etc.
thanks in advance
# imageDownloader.py
# Finds and downloads all images from any given URL recursively.
# FB - 201009094
import urllib2
from os.path import basename
import urlparse
#from BeautifulSoup import BeautifulSoup # for HTML parsing
import bs4
from bs4 import BeautifulSoup
global urlList
urlList = []
# recursively download images starting from the root URL
def downloadImages(url, level, minFileSize): # the root URL is level 0
# do not go to other websites
global website
netloc = urlparse.urlsplit(url).netloc.split('.')
if netloc[-2] + netloc[-1] != website:
return
global urlList
if url in urlList: # prevent using the same URL again
return
try:
urlContent = urllib2.urlopen(url).read()
urlList.append(url)
print url
except:
return
soup = BeautifulSoup(''.join(urlContent))
# find and download all images
imgTags = soup.findAll('img')
for imgTag in imgTags:
imgUrl = imgTag['src']
# download only the proper image files
if imgUrl.lower().endswith('.jpeg') or \
imgUrl.lower().endswith('.jpg') or \
imgUrl.lower().endswith('.gif') or \
imgUrl.lower().endswith('.png') or \
imgUrl.lower().endswith('.bmp'):
try:
imgData = urllib2.urlopen(imgUrl).read()
if len(imgData) >= minFileSize:
print " " + imgUrl
fileName = basename(urlsplit(imgUrl)[2])
output = open(fileName,'wb')
output.write(imgData)
output.close()
except:
pass
print
print
# if there are links on the webpage then recursively repeat
if level > 0:
linkTags = soup.findAll('a')
if len(linkTags) > 0:
for linkTag in linkTags:
try:
linkUrl = linkTag['href']
downloadImages(linkUrl, level - 1, minFileSize)
except:
pass
# main
rootUrl = 'http://www.leveldesigninspirationmachine.tumblr.com'
netloc = urlparse.urlsplit(rootUrl).netloc.split('.')
global website
website = netloc[-2] + netloc[-1]
downloadImages(rootUrl, 1, 50000)
As Frxstream has commented, this program creates the files in the current directory (i.e. where you run it). After running the program, run ls -l (or dir) to find the files it has created.
If it seemingly hasn't created any files, then most probably it really hasn't created any files, most probably because there was an exception which your except: pass has hidden. To see what was going wrong, replace try: ... except: pass with just ..., and rerun the program. (If you can't understand and fix that, ask a separate StackOverflow question.)
it's hard to tell without looking at the errors (+1 to turning off your try/except block so you can see the exceptions) but I do see one typo here:
fileName = basename(urlsplit(imgUrl)[2])
you didn't do "from urlparse import urlsplit" you have "import urlparse" so you need to refer to it as urlparse.urlsplit() as you have in other places, so should be like this
fileName = basename(urlparse.urlsplit(imgUrl)[2])