I've looked all around Google and its archives. There are several good articles, but none seem to help me out. So I thought I'd come here for a more specific answer.
The Objective: I want to run this code on a website to get all the picture files at once. It'll save a lot of pointing and clicking.
I've got Python 2.3.5 on a Windows 7 x64 machine. It's installed in C:\Python23.
How do I get this script to "go", so to speak?
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WOW. 35k views. Seeing as how this is top result on Google, here's a useful link I found over the years:
http://learnpythonthehardway.org/book/ex1.html
For setup, see exercise 0.
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FYI: I've got zero experience with Python. Any advice would be appreciated.
As requested, here's the code I'm using:
"""
dumpimages.py
Downloads all the images on the supplied URL, and saves them to the
specified output file ("/test/" by default)
Usage:
python dumpimages.py http://example.com/ [output]
"""
from BeautifulSoup import BeautifulSoup as bs
import urlparse
from urllib2 import urlopen
from urllib import urlretrieve
import os
import sys
def main(url, out_folder="C:\asdf\"):
"""Downloads all the images at 'url' to /test/"""
soup = bs(urlopen(url))
parsed = list(urlparse.urlparse(url))
for image in soup.findAll("img"):
print "Image: %(src)s" % image
filename = image["src"].split("/")[-1]
parsed[2] = image["src"]
outpath = os.path.join(out_folder, filename)
if image["src"].lower().startswith("http"):
urlretrieve(image["src"], outpath)
else:
urlretrieve(urlparse.urlunparse(parsed), outpath)
def _usage():
print "usage: python dumpimages.py http://example.com [outpath]"
if __name__ == "__main__":
url = sys.argv[-1]
out_folder = "/test/"
if not url.lower().startswith("http"):
out_folder = sys.argv[-1]
url = sys.argv[-2]
if not url.lower().startswith("http"):
_usage()
sys.exit(-1)
main(url, out_folder)
Since you seem to be on windows you can do this so python <filename.py>. Check that python's bin folder is in your PATH, or you can do c:\python23\bin\python <filename.py>. Python is an interpretive language and so you need the interpretor to run your file, much like you need java runtime to run a jar file.
Your command should include the url parameter as stated in the script usage comments. The main function has 2 parameters, url and out (which is set to a default value) C:\python23\python "C:\PathToYourScript\SCRIPT.py" http://yoururl.com "C:\OptionalOutput\"
Usually you can double click the .py file in Windows explorer to run it. If this doesn't work, you can create a batch file in the same directory with the following contents:
C:\python23\python YOURSCRIPTNAME.py
Then double click that batch file. Or, you can simply run that line in the command prompt while your working directory is the location of your script.
If you want to run .py files in Windows, Try installing Git bash Then download python(Required Version) from python.org and install in the main c drive folder
For me, its :
"C:\Python38"
then open Git Bash and go to the respective folder where your .py file is stored :
For me, its :
File Location : "Downloads" File Name : Train.py
So i changed my Current working Directory From "C:/User/(username)/" to "C:/User/(username)/Downloads"
then i will run the below command
" /c/Python38/python Train.py "
and it will run successfully.
But if it give the below error :
from sklearn.model_selection import train_test_split ModuleNotFoundError: No module named 'sklearn'
Then Do not panic :
and use this command :
" /c/Python38/Scripts/pip install sklearn "
and after it has installed sklearn go back and run the previous command :
" /c/Python38/python Train.py "
and it will run successfully.
!!!!HAPPY LEARNING !!!!
use IDLE Editor {You may already have it} it has interactive shell for python and it will show you execution and result.
Source: Stackoverflow.com