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  • python getelementbyid from string

    - by matthewgall
    Hey, I have the following program, that is trying to upload a file (or files) to an image upload site, however I am struggling to find out how to parse the returned HTML to grab the direct link (contained in a ). I have the code below: #!/usr/bin/python # -*- coding: utf-8 -*- import pycurl import urllib import urlparse import xml.dom.minidom import StringIO import sys import gtk import os import imghdr import locale import gettext try: import pynotify except: print "Please install pynotify." APP="Uploadir Uploader" DIR="locale" locale.setlocale(locale.LC_ALL, '') gettext.bindtextdomain(APP, DIR) gettext.textdomain(APP) _ = gettext.gettext ##STRINGS uploading = _("Uploading image to Uploadir.") oneimage = _("1 image has been successfully uploaded.") multimages = _("images have been successfully uploaded.") uploadfailed = _("Unable to upload to Uploadir.") class Uploadir: def __init__(self, args): self.images = [] self.urls = [] self.broadcasts = [] self.username="" self.password="" if len(args) == 1: return else: for file in args: if file == args[0] or file == "": continue if file.startswith("-u"): self.username = file.split("-u")[1] #print self.username continue if file.startswith("-p"): self.password = file.split("-p")[1] #print self.password continue self.type = imghdr.what(file) self.images.append(file) for file in self.images: self.upload(file) self.setClipBoard() self.broadcast(self.broadcasts) def broadcast(self, l): try: str = '\n'.join(l) n = pynotify.Notification(str) n.set_urgency(pynotify.URGENCY_LOW) n.show() except: for line in l: print line def upload(self, file): #Try to login cookie_file_name = "/tmp/uploadircookie" if ( self.username!="" and self.password!=""): print "Uploadir authentication in progress" l=pycurl.Curl() loginData = [ ("username",self.username),("password", self.password), ("login", "Login") ] l.setopt(l.URL, "http://uploadir.com/user/login") l.setopt(l.HTTPPOST, loginData) l.setopt(l.USERAGENT,"User-Agent: Uploadir (Python Image Uploader)") l.setopt(l.FOLLOWLOCATION,1) l.setopt(l.COOKIEFILE,cookie_file_name) l.setopt(l.COOKIEJAR,cookie_file_name) l.setopt(l.HEADER,1) loginDataReturnedBuffer = StringIO.StringIO() l.setopt( l.WRITEFUNCTION, loginDataReturnedBuffer.write ) if l.perform(): self.broadcasts.append("Login failed. Please check connection.") l.close() return loginDataReturned = loginDataReturnedBuffer.getvalue() l.close() #print loginDataReturned if loginDataReturned.find("<li>Your supplied username or password is invalid.</li>")!=-1: self.broadcasts.append("Uploadir authentication failed. Username/password invalid.") return else: self.broadcasts.append("Uploadir authentication successful.") #cookie = loginDataReturned.split("Set-Cookie: ")[1] #cookie = cookie.split(";",0) #print cookie c = pycurl.Curl() values = [ ("file", (c.FORM_FILE, file)) ] buf = StringIO.StringIO() c.setopt(c.URL, "http://uploadir.com/file/upload") c.setopt(c.HTTPPOST, values) c.setopt(c.COOKIEFILE, cookie_file_name) c.setopt(c.COOKIEJAR, cookie_file_name) c.setopt(c.WRITEFUNCTION, buf.write) if c.perform(): self.broadcasts.append(uploadfailed+" "+file+".") c.close() return self.result = buf.getvalue() #print self.result c.close() doc = urlparse.urlparse(self.result) self.urls.append(doc.getElementsByTagName("download")[0].childNodes[0].nodeValue) def setClipBoard(self): c = gtk.Clipboard() c.set_text('\n'.join(self.urls)) c.store() if len(self.urls) == 1: self.broadcasts.append(oneimage) elif len(self.urls) != 0: self.broadcasts.append(str(len(self.urls))+" "+multimages) if __name__ == '__main__': uploadir = Uploadir(sys.argv) Any help would be gratefully appreciated. Warm regards,

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  • Enthought Python, Sage, or others (in Unix clusters)

    - by vailen
    I am currently get access to a cluster of Unix machines, but they don't have the software I need (numpy, scipy, matplotlib, etc), and I have to install them by myself (I don't have the root permission, either, so commands like apt-get or yast doesn't work). In the worst case, I have to compile them all from source. Is there any better way to do so? I hear something about Enthought Python and Sage, but not sure what is the best way to do so. Any suggestion?

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  • Python, print delimited list

    - by Mike
    Consider this Python code for printing a list of comma separated values for element in list: print element + ",", What is the preferred method for printing such that a comma does not appear if element is the final element in the list. ex a = [1, 2, 3] for element in a print str(element) +",", output 1,2,3, desired 1,2,3

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  • Pycassa | Python

    - by MMRUser
    Is anyone having experience working with pycassa I have a doubt with it. How do I get all the keys that are stored in the database? well in this small snippet we need to give the keys in order to get the associated columns (here the keys are 'foo' and 'bar'),that is fine but my requirement is to get all the keys (only keys) at once as Python list or similar data structure. cf.multiget(['foo', 'bar']) {'foo': {'column1': 'val2'}, 'bar': {'column1': 'val3', 'column2': 'val4'}} Thanks.

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  • Should Python 2.6 on OS X deal with multiple easy-install.pth files in $PYTHONPATH?

    - by ahd
    I am running ipython from sage and also am using some packages that aren't in sage (lxml, argparse) which are installed in my home directory. I have therefore ended up with a $PYTHONPATH of $HOME/sage/local/lib/python:$HOME/lib/python Python is reading and processing the first easy-install.pth it finds ($HOME/sage/local/lib/python/site-packages/easy-install.pth) but not the second, so eggs installed in $HOME/lib/python aren't added to the path. On reading the off-the-shelf site.py, I cannot for the life of me see why it's doing this. Can someone enlighten me? Or advise how to nudge Python into reading both easy-install.pth files? Consolidating both into one .pth file is a viable workaround for now, so this question is mostly for curiosity value.

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  • python subprocess block

    - by MW
    Hello. I'm having a problem with the module subprocess. I'm running a script from python through: subprocess.Popen('./run_pythia.sh',shell=True).communicate() and sometimes it just blocks and it doesn't finish to execute the script. Before I was using .wait() instead of .communicate() but then because of this: http://dcreager.net/2009/08/06/subprocess-communicate-drawbacks/ I changed to .communicate(). Nevertheless the problem continues. Can anyone help me?

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  • Python directory list returned to Django template

    - by Shu
    Total Python newb here. I have a images directory and I need to return the names and urls of those files to a django template that I can loop through for links. I know it will be the server path, but I can modify it via JS. I've tried os.walk, but I keep getting empty results.

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  • The confusion on python encoding

    - by zhangzhong
    I retrieved the data encoded in big5 from database,and I want to send the data as email of html content, the code is like this: html += """<tr><td>""" html += """unicode(rs[0], 'big5')""" # rs[0] is data encoded in big5 I run the script, but the error raised: UnicodeDecodeError: 'ascii' codec can't decode byte...... However, I tried the code in interactive python command line, there are no errors raised, could you give me the clue?

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  • Python logger dynamic filename

    - by sharjeel
    I want to configure my Python logger in such a way so that each instance of logger should log in a file having the same name as the name of the logger itself. e.g.: log_hm = logging.getLogger('healthmonitor') log_hm.info("Testing Log") # Should log to /some/path/healthmonitor.log log_sc = logging.getLogger('scripts') log_sc.debug("Testing Scripts") # Should log to /some/path/scripts.log log_cr = logging.getLogger('cron') log_cr.info("Testing cron") # Should log to /some/path/cron.log I want to keep it generic and dont want to hardcode all kind of logger names I can have. Is that possible?

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  • Python: try statement single line

    - by Brant
    Is there a way in python to turn a try/except into a single line? something like... b = 'some variable' a = c | b #try statement goes here Where b is a declared variable and c is not... so c would throw an error and a would become b...

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  • Python sock.listen(...)

    - by Ian
    All the examples I've seen of sock.listen(5) in the python documentation suggest I should set the max backlog number to be 5. This is causing a problem for my app since I'm expecting some very high volume (many concurrent connections). I set it to 200 and haven't seen any problems on my system, but was wondering how high I can set it before it causes problems.. Anyone know?

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