SQLAlchemy在没有显式声明回滚调用的情况下，会话的未提交更改是否会自动回滚？

Question

例如，此代码：

db = SQLAlchemy()

def myfunction(a):
   #somechanges in database
   if a == 2:
     return
   db.session.commit()

myfunction(2) # there were some changes here that were not committed neither rolled back
myfunction(4) # Here the changes were committed.

我的问题是，在第二次调用中，第一次更改是否与第二次更改一起提交？在

提前谢谢

Answer 1

如果您阅读the tutorial，您将看到会话具有dirty属性：

db = SQLAlchemy()

def myfunction(a):
   #somechanges in database
   if a == 2:
     return
   db.session.commit()

myfunction(2) # there were some changes here that were not committed neither rolled back
print(db.session.dirty)
myfunction(4) # Here the changes were committed.

它告诉你什么？在

Answer 2

对myFunction(4)调用中执行的更改将覆盖对myFunction(2)调用中执行的更改。无论是否提交，更新数据都是如此。在

对于添加行和列，没有数据重写。在函数调用myFunction(4)之前提交不会产生任何影响。在