How to create a movie database starting from a list of files

Question

I have a large amount of movies on my home server (4000 circa). The files are all named Title - Subtitle (year).extension. I would like to create a database (even in excel would be fine) of all my movies. The database should contain as columns: title, subtitle (if exists), year and location of the file on the server (some movies are organized in folders by genere or actor). As of now I have a bash script that returns just a txt file containing a list of files for each hard drive (each file contains the list for each hard drive). How can I create this kind of database automatically on my home server (which is running debian)?

It would be great also to automatically retrieve other informations about the movies using some movie databases api, but I guess that this would be very complicated.

Answer 1

This is a pretty broad question and not really appropriate here (this is more of a tutorial than a quick code question), but here's some strategic advice:

• Excel will open a .csv and treat commas/new lines as cells. So
• You need to iterate, maybe recursively, over the directory(ies)
• Expand the path name—if you use a high-level language like Python, this is acheived by standard functions; then use regular expressions to parse the final bit
• Store the formatted contents of each path as rows in a list
• Print that list to a text file, joining each element by commas and each row by a new line character
• Provide said file with a .csv suffix and open it in Excel

Note that if you really want a database proper, Python again is a nice choice—SQLite is part of the standard install.

Cheers, good luck

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UPDATE: Haha, you edited the question whilst I answered. It seems like everything you need is in the file name, but if you're planning on using metadata, here's a caution. Pulling the metadata out of your files can get trickier if they've not all come from the same source; not every media type has the same metadata structure, not every application that creates the files provides the same. So the logic of getting your metadata can get messy.

Is there a reason you can't use extant programs to do this?

Finally you mention getting it on your web-server; once again deferring to Python, the capacity to make the requests of your server you need is also built into the standard package.

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Final Update

Can't help you with bash; I'm all thumbs there, and I'm no expert in Python either but your goals are pretty simple. I haven't tested this—there is probably a typo or two, consider it pseudo-code that is mostly python-ready.

# import the standard libraries you'll need
import os # https://docs.python.org/2/library/os.html
import re # https://docs.python.org/2/library/re.html

# this function will walk your directories and output a list of file paths
def getFilePaths(directory):
    file_paths = []
    for root, directories, files in os.walk(directory):
        for filename in files:
            filepath = os.path.join(root, filename)
            file_paths.append(filepath)
    return file_paths



video_file_paths = getFilePaths("path/to/video/library")
output_to_csv = [];
for video_file in video_file_paths:
    base_path, fname = os.path.split(video_file) 

     """ This is a super simple bit of regex that, provided  your files are all formatted as
     written, will parse out title, subtitle, year and file extension. If your file names
     turn out to have more exceptions than you expect (I'd be shocked if not), you may need
     to make this part more robust, either with much more savvy regex, or else some conditional
     logic—maybe a recursive try... catch loop"""
    reg_ex = re.compile("/^(.*) - (.*) \((.*)\)\.(.*)$/");

    # now apply the compiled regex to each path
    name_components = reg_ex.match(fname);

    """Each output is a row of your CSV file; .join() will join the 4 elements of the regex
    match (assuming, again, that your filenames are as clean as you claim), and then add
    the basepath, so you should be building, in this loop, a list with elements like:
    title, subtitle, year, file_extension, full path"""

    output_to_csv.append("{0},{1}".format(name_components.join(","), base_path));

#create the file, making sure the location is writeable
csv_doc = open("my_video_database.csv", "w");

# now join all the rows with line breaks and write the compiled text to the file
csv_doc.write( ouput_to_csv.join("\n") ); 

#close  your new database
csv_doc.close()