SQLAlchemy 连接 - return 一个 table 的所有列
SQLAlchemy join - return all columns of one table
我目前正在加入两个 tables returning 行,其中列在列表中:
a = Table('a', server_metadata, autoload=True)
b = Table('b', server_metadata, autoload=True)
tickers = ['1', '2']
res = b.join(a).select().where(asset.c.code.in_(tickers)).execute()
这 return 是 a
和 b
中的正确行和所有列。我怎样才能只 return table b
的所有列?
select
corresponds to the SELECT ...
part of the SQL query, and defaults to *
; to limit it to b.*
, you can give it a list containing the table b
; with this construct however you need to wrap the FROM
part into select_from
否则 SQLAlchemy 将生成一个子查询片段。因此:
res = select([b]).select_from(b.join(a).\
where(asset.c.code.in_(tickers))).execute()
我目前正在加入两个 tables returning 行,其中列在列表中:
a = Table('a', server_metadata, autoload=True)
b = Table('b', server_metadata, autoload=True)
tickers = ['1', '2']
res = b.join(a).select().where(asset.c.code.in_(tickers)).execute()
这 return 是 a
和 b
中的正确行和所有列。我怎样才能只 return table b
的所有列?
select
corresponds to the SELECT ...
part of the SQL query, and defaults to *
; to limit it to b.*
, you can give it a list containing the table b
; with this construct however you need to wrap the FROM
part into select_from
否则 SQLAlchemy 将生成一个子查询片段。因此:
res = select([b]).select_from(b.join(a).\
where(asset.c.code.in_(tickers))).execute()