Python SQLAlchemy 按降序排序？
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SQLAlchemy ORDER BY DESCENDING?
提问by AP257
How can I use ORDER BY descendingin a SQLAlchemy query like the following?
如何descending在 SQLAlchemy 查询中使用 ORDER BY ，如下所示？
This query works, but returns them in ascending order:
此查询有效，但按升序返回它们：
query = (model.Session.query(model.Entry)
.join(model.ClassificationItem)
.join(model.EnumerationValue)
.filter_by(id=c.row.id)
.order_by(model.Entry.amount) #?This row :)
)
If I try:
如果我尝试：
.order_by(desc(model.Entry.amount))
then I get: NameError: global name 'desc' is not defined.
然后我得到：NameError: global name 'desc' is not defined。
采纳答案by AP257
回答by Rick
Just as an FYI, you can also specify those things as column attributes. For instance, I might have done:
仅供参考，您还可以将这些内容指定为列属性。例如，我可能已经做了：
.order_by(model.Entry.amount.desc())
This is handy since it avoids an import, and you can use it on other places such as in a relation definition, etc.
这很方便，因为它避免了import，并且您可以在其他地方使用它，例如在关系定义等中。
For more information, you can refer this
有关更多信息，您可以参考此
回答by Radu
One other thing you might do is:
您可能会做的另一件事是：
.order_by("name desc")
This will result in: ORDER BY name desc. The disadvantage here is the explicit column name used in order by.
这将导致： ORDER BY name desc。这里的缺点是按顺序使用的显式列名。
回答by anand tripathi
You can use .desc()function in your query just like this
您可以.desc()像这样在查询中使用函数
query = (model.Session.query(model.Entry)
.join(model.ClassificationItem)
.join(model.EnumerationValue)
.filter_by(id=c.row.id)
.order_by(model.Entry.amount.desc())
)
This will order by amount in descending order or
这将按金额降序排列或
query = session.query(
model.Entry
).join(
model.ClassificationItem
).join(
model.EnumerationValue
).filter_by(
id=c.row.id
).order_by(
model.Entry.amount.desc()
)
)
Use of desc function of SQLAlchemy
SQLAlchemy 的 desc 函数的使用
from sqlalchemy import desc
query = session.query(
model.Entry
).join(
model.ClassificationItem
).join(
model.EnumerationValue
).filter_by(
id=c.row.id
).order_by(
desc(model.Entry.amount)
)
)
For official docs please use the linkor check below snippet
对于官方文档，请使用链接或检查以下代码段
sqlalchemy.sql.expression.desc(column) Produce a descending ORDER BY clause element.
e.g.:
from sqlalchemy import desc stmt = select([users_table]).order_by(desc(users_table.c.name))will produce SQL as:
SELECT id, name FROM user ORDER BY name DESCThe desc() function is a standalone version of the ColumnElement.desc() method available on all SQL expressions, e.g.:
stmt = select([users_table]).order_by(users_table.c.name.desc())Parameters column – A ColumnElement (e.g. scalar SQL expression) with which to apply the desc() operation.
See also
asc()
nullsfirst()
nullslast()
Select.order_by()
sqlalchemy.sql.expression.desc(column) 产生一个降序的 ORDER BY 子句元素。
例如：
from sqlalchemy import desc stmt = select([users_table]).order_by(desc(users_table.c.name))将生成 SQL 为：
SELECT id, name FROM user ORDER BY name DESCdesc() 函数是 ColumnElement.desc() 方法的独立版本，可用于所有 SQL 表达式，例如：
stmt = select([users_table]).order_by(users_table.c.name.desc())参数 column – 用于应用 desc() 操作的 ColumnElement（例如标量 SQL 表达式）。
也可以看看
升序（）
空优先（）
nullslast()
Select.order_by()
回答by Michael
Complementary at @Radu answer, As in SQL, you can add the table name in the parameter if you have many table with the same attribute.
补充@Radu 答案，与在 SQL 中一样，如果您有多个具有相同属性的表，则可以在参数中添加表名。
.order_by("TableName.name desc")
回答by Artem Baranov
You can try: .order_by(ClientTotal.id.desc())
你可以试试： .order_by(ClientTotal.id.desc())
session = Session()
auth_client_name = 'client3'
result_by_auth_client = session.query(ClientTotal).filter(ClientTotal.client ==
auth_client_name).order_by(ClientTotal.id.desc()).all()
for rbac in result_by_auth_client:
print(rbac.id)
session.close()
