Python 实用宝典

Question 1

How to get the code of the headers through urllib?

Question 2

The getcode() method (Added in python2.6) returns the HTTP status code that was sent with the response, or None if the URL is no HTTP URL.

>>> a=urllib.urlopen('http://www.google.com/asdfsf')
>>> a.getcode()
404
>>> a=urllib.urlopen('http://www.google.com/')
>>> a.getcode()
200

Question 3

You can use urllib2 as well:

import urllib2

req = urllib2.Request('http://www.python.org/fish.html')
try:
    resp = urllib2.urlopen(req)
except urllib2.HTTPError as e:
    if e.code == 404:
        # do something...
    else:
        # ...
except urllib2.URLError as e:
    # Not an HTTP-specific error (e.g. connection refused)
    # ...
else:
    # 200
    body = resp.read()

Note that HTTPError is a subclass of URLError which stores the HTTP status code.

Question 4

For Python 3:

import urllib.request, urllib.error

url = 'http://www.google.com/asdfsf'
try:
    conn = urllib.request.urlopen(url)
except urllib.error.HTTPError as e:
    # Return code error (e.g. 404, 501, ...)
    # ...
    print('HTTPError: {}'.format(e.code))
except urllib.error.URLError as e:
    # Not an HTTP-specific error (e.g. connection refused)
    # ...
    print('URLError: {}'.format(e.reason))
else:
    # 200
    # ...
    print('good')

Question 5

import urllib2

try:
    fileHandle = urllib2.urlopen('http://www.python.org/fish.html')
    data = fileHandle.read()
    fileHandle.close()
except urllib2.URLError, e:
    print 'you got an error with the code', e

Question 6

After using cgi.parse_qs(), how to convert the result (dictionary) back to query string? Looking for something similar to urllib.urlencode().

Question 7

Python 3

urllib.parse.urlencode(query, doseq=False, [...])

Convert a mapping object or a sequence of two-element tuples, which may contain str or bytes objects, to a percent-encoded ASCII text string.

— Python 3 urllib.parse docs

A dict is a mapping.

Legacy Python

urllib.urlencode(query[, doseq])
Convert a mapping object or a sequence of two-element tuples to a “percent-encoded” string… a series of key=value pairs separated by '&' characters…

— Python 2.7 urllib docs

Question 8

In python3, slightly different:

from urllib.parse import urlencode
urlencode({'pram1': 'foo', 'param2': 'bar'})

output: 'pram1=foo&param2=bar'

for python2 and python3 compatibility, try this:

try:
    #python2
    from urllib import urlencode
except ImportError:
    #python3
    from urllib.parse import urlencode

Question 9

You’re looking for something exactly like urllib.urlencode()!

However, when you call parse_qs() (distinct from parse_qsl()), the dictionary keys are the unique query variable names and the values are lists of values for each name.

In order to pass this information into urllib.urlencode(), you must “flatten” these lists. Here is how you can do it with a list comprehenshion of tuples:

query_pairs = [(k,v) for k,vlist in d.iteritems() for v in vlist]
urllib.urlencode(query_pairs)

Question 10

Maybe you’re looking for something like this:

def dictToQuery(d):
  query = ''
  for key in d.keys():
    query += str(key) + '=' + str(d[key]) + "&"
  return query

It takes a dictionary and convert it to a query string, just like urlencode. It’ll append a final “&” to the query string, but return query[:-1] fixes that, if it’s an issue.

Question 11

I would like to use urllib.quote(). But python (python3) is not finding the module. Suppose, I have this line of code:

print(urllib.quote("châteu", safe=''))

How do I import urllib.quote?

import urllib or import urllib.quote both give

AttributeError: 'module' object has no attribute 'quote'

What confuses me is that urllib.request is accessible via import urllib.request

Question 12

In Python 3.x, you need to import urllib.parse.quote:

>>> import urllib.parse
>>> urllib.parse.quote("châteu", safe='')
'ch%C3%A2teu'

According to Python 2.x urllib module documentation:

NOTE

The urllib module has been split into parts and renamed in Python 3 to urllib.request, urllib.parse, and urllib.error.

Question 13

If you need to handle both Python 2.x and 3.x you can catch the exception and load the alternative.

try:
    from urllib import quote  # Python 2.X
except ImportError:
    from urllib.parse import quote  # Python 3+

You could also use the python compatibility wrapper six to handle this.

from six.moves.urllib.parse import quote

Question 14

urllib went through some changes in Python3 and can now be imported from the parse submodule

>>> from urllib.parse import quote  
>>> quote('"')                      
'%22'

Question 15

This is how I handle this, without using exceptions.

import sys
if sys.version_info.major > 2:  # Python 3 or later
    from urllib.parse import quote
else:  # Python 2
    from urllib import quote

Question 16

When I try to follow the Python Wiki’s example related to URL encoding:

>>> import urllib
>>> params = urllib.urlencode({'spam': 1, 'eggs': 2, 'bacon': 0})
>>> f = urllib.urlopen("http://www.musi-cal.com/cgi-bin/query", params)
>>> print f.read()

An error is raised on the second line:

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'module' object has no attribute 'urlencode'

What am I missing?

Question 17

urllib has been split up in Python 3.

The urllib.urlencode() function is now urllib.parse.urlencode(),

the urllib.urlopen() function is now urllib.request.urlopen().

Question 18

import urllib.parse
urllib.parse.urlencode({'spam': 1, 'eggs': 2, 'bacon': 0})

Question 19

You use the Python 2 docs but write your program in Python 3.

Question 20

I am using BeautifulSoup to scrape a url and I had the following code

import urllib
import urllib2
from BeautifulSoup import BeautifulSoup

url =  "http://www.example.com/servlet/av/ResultTemplate=AVResult.html"
req = urllib2.Request(url)
response = urllib2.urlopen(req)
the_page = response.read()
soup = BeautifulSoup(the_page)
soup.findAll('td',attrs={'class':'empformbody'})

Now in the above code we can use findAll to get tags and information related to them, but I want to use xpath. Is it possible to use xpath with BeautifulSoup? If possible, can anyone please provide me an example code so that it will be more helpful?

Question 21

Nope, BeautifulSoup, by itself, does not support XPath expressions.

An alternative library, lxml, does support XPath 1.0. It has a BeautifulSoup compatible mode where it’ll try and parse broken HTML the way Soup does. However, the default lxml HTML parser does just as good a job of parsing broken HTML, and I believe is faster.

Once you’ve parsed your document into an lxml tree, you can use the .xpath() method to search for elements.

try:
    # Python 2
    from urllib2 import urlopen
except ImportError:
    from urllib.request import urlopen
from lxml import etree

url =  "http://www.example.com/servlet/av/ResultTemplate=AVResult.html"
response = urlopen(url)
htmlparser = etree.HTMLParser()
tree = etree.parse(response, htmlparser)
tree.xpath(xpathselector)

There is also a dedicated lxml.html() module with additional functionality.

Note that in the above example I passed the response object directly to lxml, as having the parser read directly from the stream is more efficient than reading the response into a large string first. To do the same with the requests library, you want to set stream=True and pass in the response.raw object after enabling transparent transport decompression:

import lxml.html
import requests

url =  "http://www.example.com/servlet/av/ResultTemplate=AVResult.html"
response = requests.get(url, stream=True)
response.raw.decode_content = True
tree = lxml.html.parse(response.raw)

Of possible interest to you is the CSS Selector support; the CSSSelector class translates CSS statements into XPath expressions, making your search for td.empformbody that much easier:

from lxml.cssselect import CSSSelector

td_empformbody = CSSSelector('td.empformbody')
for elem in td_empformbody(tree):
    # Do something with these table cells.

Coming full circle: BeautifulSoup itself does have very complete CSS selector support:

for cell in soup.select('table#foobar td.empformbody'):
    # Do something with these table cells.

Question 22

I can confirm that there is no XPath support within Beautiful Soup.

Question 23

As others have said, BeautifulSoup doesn’t have xpath support. There are probably a number of ways to get something from an xpath, including using Selenium. However, here’s a solution that works in either Python 2 or 3:

from lxml import html
import requests

page = requests.get('http://econpy.pythonanywhere.com/ex/001.html')
tree = html.fromstring(page.content)
#This will create a list of buyers:
buyers = tree.xpath('//div[@title="buyer-name"]/text()')
#This will create a list of prices
prices = tree.xpath('//span[@class="item-price"]/text()')

print('Buyers: ', buyers)
print('Prices: ', prices)

I used this as a reference.

Question 24

BeautifulSoup has a function named findNext from current element directed childern,so:

father.findNext('div',{'class':'class_value'}).findNext('div',{'id':'id_value'}).findAll('a')

Above code can imitate the following xpath:

div[class=class_value]/div[id=id_value]

Question 25

I’ve searched through their docs and it seems there is not xpath option. Also, as you can see here on a similar question on SO, the OP is asking for a translation from xpath to BeautifulSoup, so my conclusion would be – no, there is no xpath parsing available.

Question 26

when you use lxml all simple:

tree = lxml.html.fromstring(html)
i_need_element = tree.xpath('//a[@class="shared-components"]/@href')

but when use BeautifulSoup BS4 all simple too:

first remove “//” and “@”
second – add star before “=”

try this magic:

soup = BeautifulSoup(html, "lxml")
i_need_element = soup.select ('a[class*="shared-components"]')

as you see, this does not support sub-tag, so i remove “/@href” part

Question 27

Maybe you can try the following without XPath

from simplified_scrapy.simplified_doc import SimplifiedDoc 
html = '''
<html>
<body>
<div>
    <h1>Example Domain</h1>
    <p>This domain is for use in illustrative examples in documents. You may use this
    domain in literature without prior coordination or asking for permission.</p>
    <p><a href="https://www.iana.org/domains/example">More information...</a></p>
</div>
</body>
</html>
'''
# What XPath can do, so can it
doc = SimplifiedDoc(html)
# The result is the same as doc.getElementByTag('body').getElementByTag('div').getElementByTag('h1').text
print (doc.body.div.h1.text)
print (doc.div.h1.text)
print (doc.h1.text) # Shorter paths will be faster
print (doc.div.getChildren())
print (doc.div.getChildren('p'))

Question 28

from lxml import etree
from bs4 import BeautifulSoup
soup = BeautifulSoup(open('path of your localfile.html'),'html.parser')
dom = etree.HTML(str(soup))
print dom.xpath('//*[@id="BGINP01_S1"]/section/div/font/text()')

Above used the combination of Soup object with lxml and one can extract the value using xpath

Question 29

This is a pretty old thread, but there is a work-around solution now, which may not have been in BeautifulSoup at the time.

Here is an example of what I did. I use the “requests” module to read an RSS feed and get its text content in a variable called “rss_text”. With that, I run it thru BeautifulSoup, search for the xpath /rss/channel/title, and retrieve its contents. It’s not exactly XPath in all its glory (wildcards, multiple paths, etc.), but if you just have a basic path you want to locate, this works.

from bs4 import BeautifulSoup
rss_obj = BeautifulSoup(rss_text, 'xml')
cls.title = rss_obj.rss.channel.title.get_text()

Question 30

I am trying to automate download of historic stock data using python. The URL I am trying to open responds with a CSV file, but I am unable to open using urllib2. I have tried changing user agent as specified in few questions earlier, I even tried to accept response cookies, with no luck. Can you please help.

Note: The same method works for yahoo Finance.

Code:

import urllib2,cookielib

site= "http://www.nseindia.com/live_market/dynaContent/live_watch/get_quote/getHistoricalData.jsp?symbol=JPASSOCIAT&fromDate=1-JAN-2012&toDate=1-AUG-2012&datePeriod=unselected&hiddDwnld=true"

hdr = {'User-Agent':'Mozilla/5.0'}

req = urllib2.Request(site,headers=hdr)

page = urllib2.urlopen(req)

Error

File “C:\Python27\lib\urllib2.py”, line 527, in http_error_default raise HTTPError(req.get_full_url(), code, msg, hdrs, fp) urllib2.HTTPError: HTTP Error 403: Forbidden

Thanks for your assistance

Question 31

By adding a few more headers I was able to get the data:

import urllib2,cookielib

site= "http://www.nseindia.com/live_market/dynaContent/live_watch/get_quote/getHistoricalData.jsp?symbol=JPASSOCIAT&fromDate=1-JAN-2012&toDate=1-AUG-2012&datePeriod=unselected&hiddDwnld=true"
hdr = {'User-Agent': 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.11 (KHTML, like Gecko) Chrome/23.0.1271.64 Safari/537.11',
       'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8',
       'Accept-Charset': 'ISO-8859-1,utf-8;q=0.7,*;q=0.3',
       'Accept-Encoding': 'none',
       'Accept-Language': 'en-US,en;q=0.8',
       'Connection': 'keep-alive'}

req = urllib2.Request(site, headers=hdr)

try:
    page = urllib2.urlopen(req)
except urllib2.HTTPError, e:
    print e.fp.read()

content = page.read()
print content

Actually, it works with just this one additional header:

'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8',

Question 32

This will work in Python 3

import urllib.request

user_agent = 'Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.9.0.7) Gecko/2009021910 Firefox/3.0.7'

url = "http://en.wikipedia.org/wiki/List_of_TCP_and_UDP_port_numbers"
headers={'User-Agent':user_agent,} 

request=urllib.request.Request(url,None,headers) #The assembled request
response = urllib.request.urlopen(request)
data = response.read() # The data u need

Question 33

NSE website has changed and the older scripts are semi-optimum to current website. This snippet can gather daily details of security. Details include symbol, security type, previous close, open price, high price, low price, average price, traded quantity, turnover, number of trades, deliverable quantities and ratio of delivered vs traded in percentage. These conveniently presented as list of dictionary form.

Python 3.X version with requests and BeautifulSoup

from requests import get
from csv import DictReader
from bs4 import BeautifulSoup as Soup
from datetime import date
from io import StringIO 

SECURITY_NAME="3MINDIA" # Change this to get quote for another stock
START_DATE= date(2017, 1, 1) # Start date of stock quote data DD-MM-YYYY
END_DATE= date(2017, 9, 14)  # End date of stock quote data DD-MM-YYYY


BASE_URL = "https://www.nseindia.com/products/dynaContent/common/productsSymbolMapping.jsp?symbol={security}&segmentLink=3&symbolCount=1&series=ALL&dateRange=+&fromDate={start_date}&toDate={end_date}&dataType=PRICEVOLUMEDELIVERABLE"




def getquote(symbol, start, end):
    start = start.strftime("%-d-%-m-%Y")
    end = end.strftime("%-d-%-m-%Y")

    hdr = {'User-Agent': 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.11 (KHTML, like Gecko) Chrome/23.0.1271.64 Safari/537.11',
         'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8',
         'Referer': 'https://cssspritegenerator.com',
         'Accept-Charset': 'ISO-8859-1,utf-8;q=0.7,*;q=0.3',
         'Accept-Encoding': 'none',
         'Accept-Language': 'en-US,en;q=0.8',
         'Connection': 'keep-alive'}

    url = BASE_URL.format(security=symbol, start_date=start, end_date=end)
    d = get(url, headers=hdr)
    soup = Soup(d.content, 'html.parser')
    payload = soup.find('div', {'id': 'csvContentDiv'}).text.replace(':', '\n')
    csv = DictReader(StringIO(payload))
    for row in csv:
        print({k:v.strip() for k, v in row.items()})


 if __name__ == '__main__':
     getquote(SECURITY_NAME, START_DATE, END_DATE)

Besides this is relatively modular and ready to use snippet.

Question 34

I’m trying to use Python to download the HTML source code of a website but I’m receiving this error.

Traceback (most recent call last):  
    File "C:\Users\Sergio.Tapia\Documents\NetBeansProjects\DICParser\src\WebDownload.py", line 3, in <module>
     file = urllib.urlopen("http://www.python.org")
AttributeError: 'module' object has no attribute 'urlopen'

I’m following the guide here: http://www.boddie.org.uk/python/HTML.html

import urllib

file = urllib.urlopen("http://www.python.org")
s = file.read()
f.close()

#I'm guessing this would output the html source code?
print(s)

I’m using Python 3.

Question 35

This works in Python 2.x.

For Python 3 look in the docs:

import urllib.request

with urllib.request.urlopen("http://www.python.org") as url:
    s = url.read()
    # I'm guessing this would output the html source code ?
    print(s)

Question 36

A Python 2+3 compatible solution is:

import sys

if sys.version_info[0] == 3:
    from urllib.request import urlopen
else:
    # Not Python 3 - today, it is most likely to be Python 2
    # But note that this might need an update when Python 4
    # might be around one day
    from urllib import urlopen


# Your code where you can use urlopen
with urlopen("http://www.python.org") as url:
    s = url.read()

print(s)

Question 37

import urllib.request as ur
s = ur.urlopen("http://www.google.com")
sl = s.read()
print(sl)

In Python v3 the “urllib.request” is a module by itself, therefore “urllib” cannot be used here.

Question 38

To get ‘dataX = urllib.urlopen(url).read()‘ working in python3 (this would have been correct for python2) you must just change 2 little things.

1: The urllib statement itself (add the .request in the middle):

dataX = urllib.request.urlopen(url).read()

2: The import statement preceding it (change from ‘import urlib’ to:

import urllib.request

And it should work in python3 :)

Question 39

import urllib.request as ur

filehandler = ur.urlopen ('http://www.google.com')
for line in filehandler:
    print(line.strip())

Question 40

For python 3, try something like this:

import urllib.request
urllib.request.urlretrieve('http://crcv.ucf.edu/THUMOS14/UCF101/UCF101/v_YoYo_g19_c02.avi', "video_name.avi")

It will download the video to the current working directory

I got help from HERE

Question 41

Solution for python3:

from urllib.request import urlopen

url = 'http://www.python.org'
file = urlopen(url)
html = file.read()
print(html)

Question 42

Change TWO lines:

import urllib.request #line1

#Replace
urllib.urlopen("http://www.python.org")
#To
urllib.request.urlopen("http://www.python.org") #line2

If You got ERROR 403: Forbidden Error exception try this:

siteurl = "http://www.python.org"

req = urllib.request.Request(siteurl, headers={'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/80.0.3987.100 Safari/537.36'})
pageHTML = urllib.request.urlopen(req).read()

I hope your problem resolved.

Question 43

One of the possible way to do it:

import urllib
...

try:
    # Python 2
    from urllib2 import urlopen
except ImportError:
    # Python 3
    from urllib.request import urlopen

Question 44

Use six module to make you code compatible between python2 and python3

urllib.request.urlopen("<your-url>")```

Question 45

your code used in python2.x, you can use like this:

from urllib.request import urlopen
urlopen(url)

by the way, suggest another module called requests is more friendly to use, you can use pip install it, and use like this:

import requests
requests.get(url)
requests.post(url)

I thought it is easily to use, i am beginner too….hahah

Question 46

import urllib
import urllib.request
from bs4 import BeautifulSoup


with urllib.request.urlopen("http://www.newegg.com/") as url:
    s = url.read()
    print(s)
soup = BeautifulSoup(s, "html.parser")
all_tag_a = soup.find_all("a", limit=10)

for links in all_tag_a:
    #print(links.get('href'))
    print(links)

Question 47

I’m trying to scrape a website, but it gives me an error.

I’m using the following code:

import urllib.request
from bs4 import BeautifulSoup

get = urllib.request.urlopen("https://www.website.com/")
html = get.read()

soup = BeautifulSoup(html)

print(soup)

And I’m getting the following error:

File "C:\Python34\lib\encodings\cp1252.py", line 19, in encode
    return codecs.charmap_encode(input,self.errors,encoding_table)[0]
UnicodeEncodeError: 'charmap' codec can't encode characters in position 70924-70950: character maps to <undefined>

What can I do to fix this?

Question 48

I was getting the same UnicodeEncodeError when saving scraped web content to a file. To fix it I replaced this code:

with open(fname, "w") as f:
    f.write(html)

with this:

import io
with io.open(fname, "w", encoding="utf-8") as f:
    f.write(html)

Using io gives you backward compatibility with Python 2.

If you only need to support Python 3 you can use the builtin open function instead:

with open(fname, "w", encoding="utf-8") as f:
    f.write(html)

Question 49

I fixed it by adding .encode("utf-8") to soup.

That means that print(soup) becomes print(soup.encode("utf-8")).

Question 50

In Python 3.7, and running Windows 10 this worked (I am not sure whether it will work on other platforms and/or other versions of Python)

Replacing this line:

with open('filename', 'w') as f:

With this:

with open('filename', 'w', encoding='utf-8') as f:

The reason why it is working is because the encoding is changed to UTF-8 when using the file, so characters in UTF-8 are able to be converted to text, instead of returning an error when it encounters a UTF-8 character that is not suppord by the current encoding.

Question 51

While saving the response of get request, same error was thrown on Python 3.7 on window 10. The response received from the URL, encoding was UTF-8 so it is always recommended to check the encoding so same can be passed to avoid such trivial issue as it really kills lots of time in production

import requests
resp = requests.get('https://en.wikipedia.org/wiki/NIFTY_50')
print(resp.encoding)
with open ('NiftyList.txt', 'w') as f:
    f.write(resp.text)

When I added encoding=”utf-8″ with the open command it saved the file with the correct response

with open ('NiftyList.txt', 'w', encoding="utf-8") as f:
    f.write(resp.text)

Question 52

Even I faced the same issue with the encoding that occurs when you try to print it, read/write it or open it. As others mentioned above adding .encoding=”utf-8″ will help if you are trying to print it.

soup.encode(“utf-8”)

If you are trying to open scraped data and maybe write it into a file, then open the file with (……,encoding=”utf-8″)

with open(filename_csv , ‘w’, newline=”,encoding=”utf-8″) as csv_file:

Question 53

For those still getting this error, adding encode("utf-8") to soup will also fix this.

soup = BeautifulSoup(html_doc, 'html.parser').encode("utf-8")
print(soup)

Question 54

So I’m trying to make a Python script that downloads webcomics and puts them in a folder on my desktop. I’ve found a few similar programs on here that do something similar, but nothing quite like what I need. The one that I found most similar is right here (http://bytes.com/topic/python/answers/850927-problem-using-urllib-download-images). I tried using this code:

>>> import urllib
>>> image = urllib.URLopener()
>>> image.retrieve("http://www.gunnerkrigg.com//comics/00000001.jpg","00000001.jpg")
('00000001.jpg', <httplib.HTTPMessage instance at 0x1457a80>)

I then searched my computer for a file “00000001.jpg”, but all I found was the cached picture of it. I’m not even sure it saved the file to my computer. Once I understand how to get the file downloaded, I think I know how to handle the rest. Essentially just use a for loop and split the string at the ‘00000000’.’jpg’ and increment the ‘00000000’ up to the largest number, which I would have to somehow determine. Any reccomendations on the best way to do this or how to download the file correctly?

Thanks!

EDIT 6/15/10

Here is the completed script, it saves the files to any directory you choose. For some odd reason, the files weren’t downloading and they just did. Any suggestions on how to clean it up would be much appreciated. I’m currently working out how to find out many comics exist on the site so I can get just the latest one, rather than having the program quit after a certain number of exceptions are raised.

import urllib
import os

comicCounter=len(os.listdir('/file'))+1  # reads the number of files in the folder to start downloading at the next comic
errorCount=0

def download_comic(url,comicName):
    """
    download a comic in the form of

    url = http://www.example.com
    comicName = '00000000.jpg'
    """
    image=urllib.URLopener()
    image.retrieve(url,comicName)  # download comicName at URL

while comicCounter <= 1000:  # not the most elegant solution
    os.chdir('/file')  # set where files download to
        try:
        if comicCounter < 10:  # needed to break into 10^n segments because comic names are a set of zeros followed by a number
            comicNumber=str('0000000'+str(comicCounter))  # string containing the eight digit comic number
            comicName=str(comicNumber+".jpg")  # string containing the file name
            url=str("http://www.gunnerkrigg.com//comics/"+comicName)  # creates the URL for the comic
            comicCounter+=1  # increments the comic counter to go to the next comic, must be before the download in case the download raises an exception
            download_comic(url,comicName)  # uses the function defined above to download the comic
            print url
        if 10 <= comicCounter < 100:
            comicNumber=str('000000'+str(comicCounter))
            comicName=str(comicNumber+".jpg")
            url=str("http://www.gunnerkrigg.com//comics/"+comicName)
            comicCounter+=1
            download_comic(url,comicName)
            print url
        if 100 <= comicCounter < 1000:
            comicNumber=str('00000'+str(comicCounter))
            comicName=str(comicNumber+".jpg")
            url=str("http://www.gunnerkrigg.com//comics/"+comicName)
            comicCounter+=1
            download_comic(url,comicName)
            print url
        else:  # quit the program if any number outside this range shows up
            quit
    except IOError:  # urllib raises an IOError for a 404 error, when the comic doesn't exist
        errorCount+=1  # add one to the error count
        if errorCount>3:  # if more than three errors occur during downloading, quit the program
            break
        else:
            print str("comic"+ ' ' + str(comicCounter) + ' ' + "does not exist")  # otherwise say that the certain comic number doesn't exist
print "all comics are up to date"  # prints if all comics are downloaded

Question 55

Python 2

Using urllib.urlretrieve

import urllib
urllib.urlretrieve("http://www.gunnerkrigg.com//comics/00000001.jpg", "00000001.jpg")

Python 3

Using urllib.request.urlretrieve (part of Python 3’s legacy interface, works exactly the same)

import urllib.request
urllib.request.urlretrieve("http://www.gunnerkrigg.com//comics/00000001.jpg", "00000001.jpg")

Question 56

import urllib
f = open('00000001.jpg','wb')
f.write(urllib.urlopen('http://www.gunnerkrigg.com//comics/00000001.jpg').read())
f.close()

Question 57

Just for the record, using requests library.

import requests
f = open('00000001.jpg','wb')
f.write(requests.get('http://www.gunnerkrigg.com//comics/00000001.jpg').content)
f.close()

Though it should check for requests.get() error.

Question 58

For Python 3 you will need to import import urllib.request:

import urllib.request 

urllib.request.urlretrieve(url, filename)

for more info check out the link

Question 59

Python 3 version of @DiGMi’s answer:

from urllib import request
f = open('00000001.jpg', 'wb')
f.write(request.urlopen("http://www.gunnerkrigg.com/comics/00000001.jpg").read())
f.close()

Question 60

I have found this answer and I edit that in more reliable way

def download_photo(self, img_url, filename):
    try:
        image_on_web = urllib.urlopen(img_url)
        if image_on_web.headers.maintype == 'image':
            buf = image_on_web.read()
            path = os.getcwd() + DOWNLOADED_IMAGE_PATH
            file_path = "%s%s" % (path, filename)
            downloaded_image = file(file_path, "wb")
            downloaded_image.write(buf)
            downloaded_image.close()
            image_on_web.close()
        else:
            return False    
    except:
        return False
    return True

From this you never get any other resources or exceptions while downloading.

Question 61

If you know that the files are located in the same directory dir of the website site and have the following format: filename_01.jpg, …, filename_10.jpg then download all of them:

import requests

for x in range(1, 10):
    str1 = 'filename_%2.2d.jpg' % (x)
    str2 = 'http://site/dir/filename_%2.2d.jpg' % (x)

    f = open(str1, 'wb')
    f.write(requests.get(str2).content)
    f.close()

Question 62

It’s easiest to just use .read() to read the partial or entire response, then write it into a file you’ve opened in a known good location.

Question 63

Maybe you need ‘User-Agent’:

import urllib2
opener = urllib2.build_opener()
opener.addheaders = [('User-Agent', 'Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/34.0.1847.137 Safari/537.36')]
response = opener.open('http://google.com')
htmlData = response.read()
f = open('file.txt','w')
f.write(htmlData )
f.close()

Question 64

Aside from suggesting you read the docs for retrieve() carefully (http://docs.python.org/library/urllib.html#urllib.URLopener.retrieve), I would suggest actually calling read() on the content of the response, and then saving it into a file of your choosing rather than leaving it in the temporary file that retrieve creates.

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