unable to compile using jython


I'm trying to compile a python program using jython and it's throwing below error

C:\jython2.2.1>jython test.py
Traceback (innermost last):
  (no code object) at line 0
  File "test.py", line 30
                    html = html if html else download(url, user_agent).read()
SyntaxError: invalid syntax

Below is my python program. Please let me know how to resolve this.

Who is Participating?
-Richard-Connect With a Mentor Commented:
Interesting.  I tried your original program, including that embedded "if", under Python (not Jython) 2.6.4 and it had no syntax errors.  I tried it under Jython 2.5.1 and it got past that line with the "if", but gave me a syntax error saying "with will become a reserved keyword under Python 2.6".  That's a different issue, but it had no problem with that "if" construct, which surprised me.  Well, you learn something new every day.
What version of Jython are you using?  What is your operating system environment?  I'm in a Windows environment myself.

It is possible you are now having some kind of "weird" problem with the source file itself such as an unprintable character embedded in it somewhere that you don't see in your editor, or the file is sitting on top of a bad bit on your disk somewhere.  Deleting the file and creating a new one with the same contents might be beneficial.

Another suggestion I can make for finding an inexplicable problem like this is to follow the ever-tightening net principle.  Start eliminating code and see  if you still get the error.  At some point you can narrow it down to one thing you eliminate that fixes the problem, which may or may not be the actual line which the interpreter is complaining about.  When you find that one thing you will find the true problem.
sukumarrAuthor Commented:

import sys
import re
import urllib2
import urlparse
from optparse import OptionParser

# regular expression data for each website
    ('youtube.com', '%7C(.*?videoplayback.*?)%2C'),
    ('metacafe.com', '&mediaURL=(.*?)&'),
# default user agent to use when downloading
USER_AGENT = 'pytube'
# size of file to download
CHUNK_SIZE = 1024 * 1024

def scrape(url, html=None, user_agent=None, output=None):
    """Scrape video location from given url. 

    Use html instead of downloading if passed.
    Download file to output if passed.
    Return url of video if found, else None
    netloc = urlparse.urlsplit(url).netloc
    for domain, video_re in VIDEO_DATA:
        if domain in netloc:
            html = html if html else download(url, user_agent).read()
            search = re.search(video_re, html)
            if search:
                flash_url = urllib2.unquote(search.group(1))
                if output:
                    print "Downloading flash to `%s'" % output,
                    #open(output, 'wb').write(download(flash_url, user_agent).read())
                    req = download(flash_url, user_agent)
                # testing with keyword in python
                    with open(output, 'wb') as fp:
                        chunk = True
                        while chunk:
                            chunk = req.read(CHUNK_SIZE)
                            if chunk:
                                print '.',
                return flash_url
                raise PyTubeException('Failed to locate video regular expression in downloaded HTML')
    raise PyTubeException('URL did not match available domains')

def download(url, user_agent=None):
    """Download url and return data
    headers = {'User-Agent' : user_agent}
    req = urllib2.Request(url, None, headers)
    return urllib2.urlopen(req)

class PyTubeException(Exception):

if __name__ == '__main__':
    # parse command line options
    parser = OptionParser(usage='usage: %prog, [-o <file.flv> -a <user_agent> -s -h] url')
    parser.add_option('-o', '--output', dest='output', help='Output file to download flash file to. If this is not specified file will not be downloaded.')
    parser.add_option('-s', '--sites', action='store_true', default=False, dest='sites', help='Display sites that pytube supports, then quit.')
    parser.add_option('-a', '--agent', dest='user_agent', default=USER_AGENT, help='Set user-agent for downloads.') 
    options, args = parser.parse_args()
    if options.sites:
        print '\n'.join(domain for (domain, reg) in VIDEO_DATA)
        if args:
            flash_url = scrape(args[0], user_agent=options.user_agent, output=options.output)
            if flash_url:
                print flash_url
            print 'Need to pass the url of the video you want to download'

Open in new window

I'm unaware of Python or Jython supporting an embeeded if statement such as that line contains.  I believe the intention of the code can be realized by the following:

if not html:
    html = download(url, user_agent).read()

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To elaborate somewhat, I recognize that the construct you attempted to use is similar to a Java construct which would compile just fine in Java.  You seem to be operating on the assumption that Jython is somehow "like" Java.  This is actually pretty much untrue.  Jython is an implementation of Python and syntactically is pure Python. However, Jython is implemented in Java, which allows it to integrate seamlessly with Java libraries and classes, to run under the Java virtual machine, and to be used in any Java environment.  But don't go trying to use Java syntax, it won't like it!
sukumarrAuthor Commented:

I did as you said but it's throwing error as below

C:\jython2.2.1>jython test.py
Traceback (innermost last):
  (no code object) at line 0
  File "test.py", line 30
        if not html:
SyntaxError: invalid syntax
sukumarrAuthor Commented:
Perfect...and after copying my own code to notepad(Windows cista) and re-run again it worked. Also richard, i had a last question.

After running will Jython create some kind of class file like for ex: if we run test.py..will it generate test.class ?
It is possible to make Jython do that, but it is not straigtforwards.  Here is a tutorial:  http://wiki.python.org/jython/JythonMonthly/Articles/September2006/1
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