Python - How do you run a .py file?

Question

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)

Answer 1

On windows platform, you have 2 choices:

1. In a command line terminal, type

   c:\python23\python xxxx.py

2. Open the python editor IDLE from the menu, and open xxxx.py, then press F5 to run it.

For your posted code, the error is at this line:

def main(url, out_folder="C:\asdf\"): 

It should be:

def main(url, out_folder="C:\\asdf\\"): 

Answer 2

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.

Answer 3

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.

Answer 4

use IDLE Editor {You may already have it} it has interactive shell for python and it will show you execution and result.