Splitting a query string on a & char to get different arguments; Python:

Question

I'm looking to split the current query string on a '&' char so I can get different query arguments. From those arguments I am looking to put them into different files, i.e. p_file.txt, blog_file.txt, portfolio_file.txt, etc. I have been stuck trying to split a list of queries but it is not possible. I am open for help.

def parse_file():
    # Open the file for reading
    infile = open("URLlist.txt", 'r')
    # Read every single line of the file into an array of lines
    lines = infile.readlines()

    # For every line in the array of lines, do something with that line
    for line in lines:
        # The lines we get back from readlines will have a newline
        # character appended.  So, let's strip that out as we parse
        # the URL from the line into its components
        line = line.strip()
        url = urlparse(line)
        # If the url has a query component...(ie. url.query)
        if url.query:
            # ...then print it out!  We need to strip the trailing newline
            # character from the url query, because urlparse doesn't do that
            # for us.  
            queryvars = url.query
            print queryvars
            #for q in queryvars:
                 #print q
       parse_file()

Yes! It is splitting upon the '&' char. How would I be able to put whatever is split into different files? ex: p=49 -> p_file.txt, attachment_id = 32 -> attachment_id_file.txt — Michael Jerzy Regdosz
– Michael Jerzy Regdosz, Commented Oct 16, 2013 at 18:35
@kindall's suggestion to use parse_qs is probably better, actually. That will give you a dictionary, e.g. { 'p': ['49'], 'attachment_id': ['32'] }. You can then iterate over that with e.g. for key in queryvars: filename = key + '_file.txt' [...]. — Jess
– Jess, Commented Oct 16, 2013 at 18:44
so: url = parse_qsl(urlparse(line)[4]) for key in url: p_file = key + 'p_file.txt' — Michael Jerzy Regdosz
– Michael Jerzy Regdosz, Commented Oct 16, 2013 at 19:08

kindall · Accepted Answer · 2013-10-16 18:28:27Z

1

I expect you want urlparse.parse_qs.

answered Oct 16, 2013 at 18:28

kindall

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