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okay, let's dive into how to upload a large file using python 3 with only built-in modules. we'll cover the concepts, techniques, and code examples you'll need to implement this effectively.

*understanding the challenges and approach*

uploading large files directly into memory can be incredibly inefficient and lead to memory errors (especially if you're dealing with files larger than your available ram). therefore, the core principle is to upload the file in chunks. this involves:

1. *reading the file in chunks:* you'll read the file into smaller, manageable segments (chunks).
2. *encoding the chunks:* we'll use `multipart/form-data` to encode each chunk. this is a standard way of sending files via http, breaking the file into parts and allowing metadata to be associated with each part.
3. *sending the chunks:* you'll send each chunk to the server using http post requests.
4. *server-side handling:* (we'll discuss this conceptually, as the server-side implementation depends on your server technology). the server will need to receive each chunk and reassemble them into the complete file.

*modules used*

**`http.client`**: for making http requests to the server.
**`urllib.parse`**: for handling urls (parsing, encoding data).
**`os`**: for file operations (getting file size, reading files).
**`mimetypes`**: to figure out the file's mime type, helps the server identify the content.

*code breakdown*

here's a python script that encapsulates the upload logic:



*explanation*

1. *`upload_file(host, url, filepath, chunk_size)`:*
takes the server's hostname, the upload url, the file path, and the chunk size as arguments.

2. *`filename = os.path.basename(filepath)`:* extracts the file name from the full path.

3. **`mimetype = mimetypes.guess_type(filepath)[0] or 'application/octet-stream'`**: guesses the file's mime type based on its extension. if it can't determine the type, it defaults to `application/octet-s ...

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