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我有其经由以下命令创建和PostgreSQL中填充的表：Django原子选择更新不锁定表的递归调用？
CREATE TABLE my_lock (
id integer,
CONSTRAINT id_pkey PRIMARY KEY (id)
) ;
INSERT INTO my_lock VALUES (1) ;
INSERT INTO my_lock VALUES (2) ;
该表由以下的Django模型
from django.db import models
from django.db import transaction
class MyLock(models.Model):
class Meta(object):
db_table = 'my_lock'
接着表示，我有以下几种方法：
from contextlib import contextmanager
@contextmanager
def acquire_lock():
with transaction.atomic():
lock = MyLock.objects.select_for_update().filter(id=1).first()
yield lock
def first_method():
print "In first method"
with acquire_lock():
print "Lock acquired in first_method()"
second_method()
def second_method():
print "In second method"
first_method()
acquire_lock()方法是一个Python生成器，它运行一个SELECT FOR UPDATE在事务中查询。这应该对id = 1的行进行锁定，并且由于调用yield lock时事务未完成，请持续保持该锁定。
因此，如果我们调用first_method()，以下输出应该打印：
In first method
Lock acquired in first_method()
In second method
In first method
然而，在现实中上调用first_method()，下面会打印：
In first method
Lock acquired in first_method()
In second method
In first method
Lock acquired in first_method()
In second method
In first method
Lock acquired in first_method()
In second method
In first method
Lock acquired in first_method()
In second method
（这一直持续到RuntimeError: maximum recursion depth exceeded）
我在这里错过了一些东西。这怎么会发生？ PostgreSQL中的行锁如何被多次获取？
编辑：
如果我改变first_method()到：
def first_method():
print "In first method"
with acquire_lock():
print "Lock acquired in first_method()"
i = 1
while True:
i = i + 1
i = i - 1
和，现在就从两个不同终端（或壳），first_method() 其中一个打印如下：
In first method
Lock acquired in first_method()
第二个打印以下内容：
In first method
因此，锁在这种情况下起作用，但不能递归地工作。