Python SQLAlchemy 将 SELECT 查询结果转换为字典列表
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SQLAlchemy convert SELECT query result to a list of dicts
提问by salamey
When I was using session.query, I was able to convert the result to a list of dicts :
当我使用 session.query 时，我能够将结果转换为 dicts 列表：
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 :
但是现在我必须使用该SELECT()操作，对于每一行结果，我如何将结果转换为看起来像这样的字典：
[{'Row1column1Name' : 'Row1olumn1Value', 'Row1column2Name' :'Row1Column2Value'},{'Row2column1Name' : 'Row2olumn1Value', 'Row2column2Name' : 'Row2Column2Value'},etc....].
[{'Row1column1Name' : 'Row1olumn1Value', 'Row1column2Name' :'Row1Column2Value'},{'Row2column1Name' : 'Row2olumn1Value', 'Row2column2Name' : 'Row2Column2Value'},etc....].
This is my SELECT() code :
这是我的 SELECT() 代码：
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 :
结果是：
[('value1', 'value2', 'value3', 'value4'),(.....),etc for each row]
I'm new to Python, any help would be appreciated.
我是 Python 新手，任何帮助将不胜感激。
回答by fanti
This seems to be a RowProxy object. Try:
这似乎是一个 RowProxy 对象。尝试：
row = dict(zip(row.keys(), row))
回答by Carl
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:
您可以将选择结果中的每一行类型转换为字典或元组。您所看到的是默认行为，即将每一行表示为一个元组。要将类型转换为 dict，请像这样修改您的代码：
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.
如果您需要一次处理一行结果，这很有效。
If you're happy to put all rows into a list in one go, list comprehension is a bit neater:
如果您很乐意一次性将所有行放入一个列表中，那么列表理解会更简洁一些：
select = select([table1, table2]).where(all_filters)
res = conn.execute(select)
resultset = [dict(row) for row in res]
print resultset
回答by Hymanotonye
For the first query its better to use this approach for sqlalchemy KeyedTuple:
对于第一个查询，最好将这种方法用于 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
或者
result_dict = map(lambda obj: dict(zip(obj.keys(), obj)), my_query)
For ResultProxy as earlier mentioned:
对于前面提到的 ResultProxy：
result_dict = dict(zip(row.keys(), row))
回答by Nick Throckmorton
Building off of Fanti's answer, if you use a List Comprehension, you can produce the list of dicts all in one row. Here resultsis the result of my query.
根据 Fanti 的答案，如果您使用 List Comprehension，则可以在一行中生成所有 dicts 列表。这results是我的查询结果。
records = [dict(zip(row.keys(), row)) for row in results]
