ΘΕΜΑ: Άντληση Δεδομένων από το Internet.

def remove_chrs(s):
    '''
        Strips white space and new line characters
        from the string.
 
        @author: Konstas Marmatakis
        License: GNU/GPL v3.0
 
        Returns String
 
    '''
    splited = s.split()
    nkey = ''
    try:
        for i in splited:
            nkey = nkey + ' ' + i.strip()
    except IndexError:
        nkey = splited[0].strip()
    return nkey
 
 
def listdoy():
    '''
        Takes the kodes form Tax Offices in the
        http://www.gsis.gr web site
 
        @author: Konstas Marmatakis
        License: GNU/GPL v3.0
 
        Returns dictionary
 
    '''
    import re
    import urllib.request
 
    # The pattern
    ptrn = r'<div.*>(\d+)</div>.*\s*.*<div.*>([\s*\w+\.?.*\'?\s*]+).*</div>'
 
    # Read the website
    f = urllib.request.urlopen('http://www.gsis.gr/ggps/doy/kodikoi_doy.html')
    periexomeno = str(f.read().decode('iso-8859-7'))
 
    # Special for the kodikoi_doy.html we remove the <span> tag 
    periexomeno1 = periexomeno.replace(r"<span style='mso-spacerun:yes'>", '')
    periexomeno2 = periexomeno1.replace('</span>', '')
 
    # We extract kodes and Tax offices
    # doy is a tuple
    doy = re.findall(ptrn, periexomeno2)
 
 
    doy_key = {}
 
    for num, txt in doy:
        doy_key[num] = remove_chrs(txt)
 
    return doy_key
 
if __name__ == '__main__':
    x = listdoy()
    for k in sorted(x.keys()):
        print(k + " " + x[k])

# The pattern
    ptrn = r'<div.*>(\d+)</div>.*\s*.*<div.*>([\s*\w+\.?.*\'?\s*]+).*</div>'