Python convert a Tuple to CSV string-CodePudding

I'm trying to implement a REST api to returns records from a SQL database in CSV format. I would like to use the import csv to do the conversion and I would like to use a generator so that I can stream the results.

This is what I have tried.

def get_data(query) -> Generator[str, None, None]:
    with pyodbc.connect(connStr) as conn:
        with conn.cursor() as cursor:
            cursor.execute(query)
            while 1:
                row = cursor.fetchone()
                if not row: break
                data = io.StringIO()
                csv.writer(data).writerow(row)
                yield data.getvalue()

This works, but I don't like it. It seems too verbose and it creates so many temporary StringIO and writer objects!! Please tell me a better way.

CodePudding user response：

I suppose I could reuse the StringIO using a seek...

def get_data(query) -> Generator[str, None, None]:
    data = io.StringIO()
    writer = csv.writer(data)
    with pyodbc.connect(connStr) as conn:
        with conn.cursor() as cursor:
            cursor.execute(query)
            while 1:
                row = cursor.fetchone()
                if not row: break
                data.seek(0)
                writer.writerow(row)
                yield data.getvalue()

CodePudding user response：

If the goal is just return/yield a resultset record values (gathered as tuple) in csv format you don't need overcomplicating with io.StringIO() and csv.writer:

def get_data(query) -> Generator[str, None, None]:
    with pyodbc.connect(connStr) as conn:
        with conn.cursor() as cursor:
            cursor.execute(query)
            while True:
                row = cursor.fetchone()
                if not row: break
                yield ', '.join(map(str, (row)))