use*_*574 22 python sqlalchemy
我想做这样的事情:
select username, userid, 'user' as new_column from users_table.
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可以使用sqlalchemy选择表的列,如下所示:
query = select([users_table.c.username, users_table.c.userid])
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我该怎么做选择x是col_x在SQLAlchemy的查询?
pyl*_*ver 45
用这个: users_table.c.userid.label('NewColumn')
即
query = select([users_table.c.username, users_table.c.userid.label('NewColumn')])
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评估为:
SELECT username , userid as NewColumn From MyTable;
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也许literal_column?
query = select([users_table.c.username, users_table.c.userid, literal_column("user", type_=Unicode).label('new_column')])
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参见https://docs.sqlalchemy.org/zh/13/core/sqlelement.html#sqlalchemy.sql.expression.literal_column
编辑:实际上我应该说“文字”:
query = select([users_table.c.username, users_table.c.userid, literal("user", type_=Unicode).label('new_column')])
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Rad*_*ski -2
请参阅alias()此处的文档:https://docs.sqlalchemy.org/en/14/core/selectable.html ?highlight=alias#sqlalchemy.sql.expression.Alias
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