Python——如何运行.py文件?

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.

我查看了谷歌和它的档案。有几篇好文章，但似乎没有一篇对我有帮助。所以我想我来这里是为了得到一个更具体的答案。

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.

目标:我想在一个网站上运行这个代码，一次获取所有的图片文件。它将节省大量的指向和点击。

I've got Python 2.3.5 on a Windows 7 x64 machine. It's installed in C:\Python23.

我在Windows 7 x64机上安装了Python 2.3.5。它是安装在C:\ Python23。

How do I get this script to "go", so to speak?

我怎样才能让这个脚本“go”呢?

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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:

哇。35 k的观点。这是我多年来发现的一个有用的链接。

http://learnpythonthehardway.org/book/ex1.html

For setup, see exercise 0.

对于设置，请参见练习0。

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FYI: I've got zero experience with Python. Any advice would be appreciated.

我对Python没有任何经验。如有任何建议，我们将不胜感激。

As requested, here's the code I'm using:

按照要求，以下是我使用的代码:

"""
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)

5 个解决方案

#1

On windows platform, you have 2 choices:

在windows平台上，你有两个选择:

In a command line terminal, type

在命令行终端中，输入

c:\python23\python xxxx.py

c:\ python23 \ python xxxx.py
Open the python editor IDLE from the menu, and open xxxx.py, then press F5 to run it.

从菜单中打开空闲的python编辑器，并打开xxxx。py，然后按F5运行它。

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

对于您发布的代码，错误如下:

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

It should be:

应该是:

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

#2

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.

由于您似乎在windows上，您可以这样做，所以python 。检查python的bin文件夹是否在您的路径中，或者您可以执行c:\python23\bin\python 。Python是一种解释性语言，所以您需要解释器来运行文件，就像您需要java运行时来运行jar文件一样。

#3

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

使用空闲编辑器{您可能已经有它}它有python的交互shell，它将显示您的执行和结果。

#4

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:

通常您可以双击Windows资源管理器中的.py文件来运行它。如果这不起作用，您可以在相同的目录中创建一个具有以下内容的批处理文件:

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.

然后双击该批处理文件。或者，您可以在命令提示符中运行这一行，而您的工作目录是脚本的位置。

#5

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\"

您的命令应该包括url参数，如脚本使用注释中所述。主函数有两个参数，url和out(设置为默认值)C:\python23\python“C:\PathToYourScript脚本”。py " http://yoururl.com " C:\ OptionalOutput \”

#1