[英]Convert raw SQL to SQLAlchemy query
SELECT *
FROM product_stocks
WHERE detected_date = (
SELECT MAX(detected_date)
FROM product_stocks
WHERE id = 18865
)
AND id = 18865;
将其转换为 SQLAlchemy 查询字符串时遇到很多麻烦。 最有效的方法是什么?
您可以使用from_statement
执行原始 SQL-Query 并在 SQL-Alchemy 对象中获取它。 当编写普通 SQL 比编写 SQLAlchemy 语法更容易时,这会有所帮助。
Session.query(YourClass).from_statement(text('''SELECT * FROM product_stocks
WHERE detected_date = (SELECT MAX(detected_date) FROM product_stocks WHERE id = 18865)
AND id = 18865;''')).all()
下面将重新创建您要求的 SQL:
_id = 18865
T = aliased(ProductStock, name="T")
T1 = aliased(ProductStock, name="T1")
subquery = (
session.query(func.max(T1.detected_date).label("detected_date"))
.filter(T1.id == _id)
# .filter(T1.id == T.id) # @note: i prefer this one to the line above
.as_scalar()
)
qry = (
session.query(T)
.filter(T.detected_date == subquery)
.filter(T.id == _id)
)
这是实现您想要的最有效的方法吗? - 我不太确定,但没有足够的信息
使用核心 SQLAlchemy 1.4/2.0:
from sqlalchemy import text, select, column
sql = 'SELECT foo FROM bar'
sql = text(sql)
sql = sql.columns(column('foo')) # This let's it be used as a subquery
sel = select(sql.selected_columns.foo).select_from(sql.subquery())
joined = sel.outerjoin(baz_t, baz_t.foo==sel.c.foo)
final = select(sel.c.foo).select_from(joined)
使用核心 SQLAlchemy < 1.4:
sql = 'SELECT foo FROM bar' sql = text(sql) sql = sql.columns() # This let's it be used as a subquery sel = select(['foo']).select_from(sql) # I needed this for a complex query or else columns would be ambiguous sel = sel.alias('sel') joined = sel.outerjoin(baz_t, baz_t.foo==sel.c.foo) final = select([sel.c.foo]).select_from(joined)
请注意, columns()
是必需的,如果查询很复杂,则alias()
很有帮助。
以下文本文档很有帮助。
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