I copy some Python code in order to download data from a website. Here is my specific website:
https://www.codot.gov/business/bidding/bid-tab-archives/bid-tabs-2017-1
Here is the code which I copied:
import requests
from bs4 import BeautifulSoup
def _getUrls_(res):
hrefs = []
soup = BeautifulSoup(res.text, 'lxml')
main_content = soup.find('div',{'id' : 'content-core'})
table = main_content.find("table")
for a in table.findAll('a', href=True):
hrefs.append(a['href'])
return(hrefs)
bidurl = 'https://www.codot.gov/business/bidding/bid-tab-archives/bid-tabs-2017-1'
r = requests.get(bidurl)
hrefs = _getUrls_(r)
def _getPdfs_(hrefs, basedir):
for i in range(len(hrefs)):
print(hrefs[i])
respdf = requests.get(hrefs[i])
pdffile = basedir + "/pdf_dot/" + hrefs[i].split("/")[-1] + ".pdf"
try:
with open(pdffile, 'wb') as p:
p.write(respdf.content)
p.close()
except FileNotFoundError:
print("No PDF produced")
basedir= "/Users/ABC/Desktop"
_getPdfs_(hrefs, basedir)
The code runs successfully, but it did not download anything at all, even though there is no Filenotfounderror obviously.
I tried the following two URLs:
https://www.codot.gov/business/bidding/bid-tab-archives/bid-tabs-2017/aqc-088a-035-20360
https://www.codot.gov/business/bidding/bid-tab-archives/bid-tabs-2017/aqc-r100-258-21125
However both of these URLs return >>> No PDF produced.
The thing is that the code worked and downloaded successfully for other people, but not me.
Your code works I just tested. You need to make sure the basedir exists, you want to add this to your code:
if not os.path.exists(basedir):
os.makedirs(basedir)
I used this exact (indented) code but replaced the basedir with my own dir and it worked only after I made sure that the path actually exists. This code does not create the folder in case it does not exist.
As others have pointed out, you need to create basedir beforehand. The user running the script may not have the directory created. Make sure you insert this code at the beginning of the script, before the main logic.
Additionally, hardcoding the base directory might not be a good idea when transferring the script to different systems. It would be preferable to use the users %USERPROFILE% enviorment variable:
from os import envioron
basedir= join(environ["USERPROFILE"], "Desktop", "pdf_dot")
Which would be the same as C:\Users\blah\Desktop\pdf_dot.
However, the above enivorment variable only works for Windows. If you want it to work Linux, you will have to use os.environ["HOME"] instead.
If you need to transfer between both systems, then you can use os.name:
from os import name
from os import environ
# Windows
if name == 'nt':
basedir= join(environ["USERPROFILE"], "Desktop", "pdf_dot")
# Linux
elif name == 'posix':
basedir = join(environ["HOME"], "Desktop", "pdf_dot")
You don't need to specify the directory or create any folder manually. All you need do is run the following script. When the execution is done, you should get a folder named pdf_dot in your desktop containing the pdf files you wish to grab.
import requests
from bs4 import BeautifulSoup
import os
URL = 'https://www.codot.gov/business/bidding/bid-tab-archives/bid-tabs-2017-1'
dirf = os.environ['USERPROFILE'] + '\Desktop\pdf_dot'
if not os.path.exists(dirf):os.makedirs(dirf)
os.chdir(dirf)
res = requests.get(URL)
soup = BeautifulSoup(res.text, 'lxml')
pdflinks = [itemlink['href'] for itemlink in soup.find_all("a",{"data-linktype":"internal"}) if "reject" not in itemlink['href']]
for pdflink in pdflinks:
filename = f'{pdflink.split("/")[-1]}{".pdf"}'
with open(filename, 'wb') as f:
f.write(requests.get(pdflink).content)
Related
I'm a python beginner and I want my code to open a local HTML page with parameters (e.g. /help/index.html#1). Right now I have the below code:
def openHelp(self, helpid):
import subprocess
import webbrowser
import sys, os
directory = os.getcwd()
if sys.platform == 'darwin': # For macOS
url = 'file://' + directory + '/help/index.html' + '#{}'.format(str(helpid))
webbrowser.get().open(url)
else:
url = 'help/index.html' + '#{}'.format(str(helpid))
webbrowser.open(url)
The code opens the webpage, however without the parameters (#helpid). Where did I make a mistake? Thanks in advance!
Your code looked fine to me. I tried it and it worked. You have to call it with openHelp("",1). The #helpid parameter was added correctly. Make sure it is a number.
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])
I have a task to downoad McAfee virus definition file daily from download site locally. The name of the file changes every day. The path to download site is "http://download.nai.com/products/licensed/superdat/english/intel/7612xdat.exe"
This ID 7612 will change every day for something different hence I can't hard code it. I have to find a way to either list file name before providing it as argument or etc.
On stackoverflow site I found someone's script which will work for me if someone could advice how to handle changing file name.
Here is the script that I'm going to use:
def download(url):
"""Copy the contents of a file from a given URL
to a local file.
"""
import urllib
webFile = urllib.urlopen(url)
localFile = open(url.split('/')[-1], 'w')
localFile.write(webFile.read())
webFile.close()
localFile.close()
if __name__ == '__main__':
import sys
if len(sys.argv) == 2:
try:
download(sys.argv[1])
except IOError:
print 'Filename not found.'
else:
import os
print 'usage: %s http://server.com/path/to/filename' % os.path.basename(sys.argv[0])
Could someone advice me?
Thanks in advance
Its a two-step process. First scrape the index page looking for files. Second, grab the latest and download
import urllib
import lxml.html
import os
import shutil
# index page
pattern_files_url = "http://download.nai.com/products/licensed/superdat/english/intel"
# relative url references based here
pattern_files_base = '/'.join(pattern_files_url.split('/')[:-1])
# scrape the index page for latest file list
doc = lxml.html.parse(pattern_files_url)
pattern_files = [ref for ref in doc.xpath("//a/#href") if ref.endswith('xdat.exe')]
if pattern_files:
pattern_files.sort()
newest = pattern_files[-1]
local_name = newest.split('/')[-1]
# grab it if we don't already have it
if not os.path.exists(local_name):
url = pattern_files_base + '/' + newest
print("downloading %s to %s" % (url, local_name))
remote = urllib.urlopen(url)
print dir(remote)
with open(local_name, 'w') as local:
shutil.copyfileobj(remote, local, length=65536)
I'm developing an application that runs on an Apache server with Django framework. My current script works fine when it runs on the local desktop (without Django). The script downloads all the images from a website to a folder on the desktop. However, when I run the script on the server a file object is just create by Django that apparently has something in it (should be google's logo), however, I can't open up the file. I also create an html file, updated image link locations, but the html file gets created fine, I'm assuming because it's all text, maybe? I believe I may have to use a file wrapper somewhere, but I'm not sure. Any help is appreciated, below is my code, Thanks!
from django.http import HttpResponse
from bs4 import BeautifulSoup as bsoup
import urlparse
from urllib2 import urlopen
from urllib import urlretrieve
import os
import sys
import zipfile
from django.core.servers.basehttp import FileWrapper
def getdata(request):
out = 'C:\Users\user\Desktop\images'
if request.GET.get('q'):
#url = str(request.GET['q'])
url = "http://google.com"
soup = bsoup(urlopen(url))
parsedURL = list(urlparse.urlparse(url))
for image in soup.findAll("img"):
print "Old Image Path: %(src)s" % image
#Get file name
filename = image["src"].split("/")[-1]
#Get full path name if url has to be parsed
parsedURL[2] = image["src"]
image["src"] = '%s\%s' % (out,filename)
print 'New Path: %s' % image["src"]
# print image
outpath = os.path.join(out, filename)
#retrieve images
if image["src"].lower().startswith("http"):
urlretrieve(image["src"], outpath)
else:
urlretrieve(urlparse.urlunparse(parsedURL), out) #Constructs URL from tuple (parsedURL)
#Create HTML File and writes to it to check output (stored in same directory).
html = soup.prettify("utf-8")
with open("output.html", "wb") as file:
file.write(html)
else:
url = 'You submitted nothing!'
return HttpResponse(url)
My problem had to do with storing the files on the desktop. I stored the files in the DJango workspace folder, changed the paths, and it worked for me.
I have a list of links, which is stored in a file. I would like to open all the links in my browsers via some script instead of manually copy-pasting every items.
For example, OS: MAC OS X; Browser: Chrome; Script: Python (prefered)
Take a look at the webbrowser module.
import webbrowser
urls = ['http://www.google.com', 'http://www.reddit.com', 'http://stackoverflow.com']
b = webbrowser.get('firefox')
for url in urls:
b.open(url)
P.S.: support for Chrome has been included in the version 3.3, but Python 3.3 is still a release candidate.
Since you're on a mac, you can just use the subprocess module to call open http://link1 http://link2 http://link3. For example:
from subprocess import call
call(["open","http://www.google.com", "http://www.stackoverflow.com"])
Note that this will just open your default browser; however, you could simply replace the open command with the specific browser's command to choose your browser.
Here's a full example for files of the general format
alink
http://anotherlink
(etc.)
from subprocess import call
import re
import sys
links = []
filename = 'test'
try:
with open(filename) as linkListFile:
for line in linkListFile:
link = line.strip()
if link != '':
if re.match('http://.+|https://.+|ftp://.+|file://.+',link.lower()):
links.append(link)
else:
links.append('http://' + link)
except IOError:
print 'Failed to open the file "%s".\nExiting.'
sys.exit()
print links
call(["open"]+links)