SQLAlchemy 将 SELECT 查询结果转换为字典列表

Question

When I was using session.query, I was able to convert the result to a list of dicts :

my_query = session.query(table1,table2).filter(all_filters)
result_dict = [u.__dict__ for u in my_query.all()]

But now that I have to work with the SELECT() operation, how can I convert the results to a dict that looks like, for every row result :

[{'Row1column1Name' : 'Row1olumn1Value', 'Row1column2Name' :'Row1Column2Value'},{'Row2column1Name' : 'Row2olumn1Value', 'Row2column2Name' : 'Row2Column2Value'},etc....] .

This is my SELECT() code :

select = select([table1,table2]).where(all_filters)
res = conn.execute(select)
row = res.fetchone() #I have to use fetchone() because the query returns lots of rows
resultset=[]
while row is not None:
    row = res.fetchone()
    resultset.append(row)

print resultset

The result is :

[('value1', 'value2', 'value3', 'value4'),(.....),etc for each row]

I'm new to Python, any help would be appreciated.

Answer 1

This seems to be a RowProxy object. Try:

row = dict(zip(row.keys(), row))

Answer 2

You can typecast each row from a select result as either a dict or a tuple. What you've been seeing is the default behaviour, which is to represent the each row as a tuple. To typecast to a dict, modify your code like this:

select = select([table1, table2]).where(all_filters)
res = conn.execute(select)
resultset = []
for row in res:
    resultset.append(dict(row))
print resultset

This works nicely if you need to process the result one row at a time.

If you're happy to put all rows into a list in one go, list comprehension is a bit neater:

select = select([table1, table2]).where(all_filters)
res = conn.execute(select)
resultset = [dict(row) for row in res]
print resultset

Answer 3

For the first query its better to use this approach for sqlalchemy KeyedTuple:

# Convert an instance of `sqlalchemy.util._collections.KeyedTuple`
# to a dictionary
my_query = session.query(table1,table2).filter(all_filters)
result_dict = map(lambda q: q._asdict(), my_query)

OR

result_dict = map(lambda obj: dict(zip(obj.keys(), obj)), my_query)

For ResultProxy as earlier mentioned:

result_dict = dict(zip(row.keys(), row))

Answer 4

Building off of Fanti's answer, if you use a List Comprehension, you can produce the list of dicts all in one row. Here results is the result of my query.

records = [dict(zip(row.keys(), row)) for row in results]

Answer 5

Make a function -

def resultToDict(result):
    ds = []
    for rows in result:
        d = {}
        for row in rows:
            for col in row.__table__.columns:
                d[col.name] = str(getattr(row, col.name))
        ds.append(d)
    return ds

Execute like this -

resultToDict(my_query.all())

Result -

[
{'Row1column1Name' : 'Row1column1Value', 'Row1column2Name' : 'Row1Column2Value'},
{'Row2column1Name' : 'Row2column1Value', 'Row2column2Name' : 'Row2Column2Value'},
{'Row3column1Name' : 'Row3column1Value', 'Row3column2Name' : 'Row3Column2Value'},
etc....
]