在 Python 和 SQLite 上运行单元测试时,我看到了这种熟悉的景象:
SQLite objects created in a thread can only be used in that same thread.
测试运行良好,但错误仍在打印出来。我假设 SQLite 对象以某种方式泄漏到后台线程。
有没有一种简单的方法来代替逐段剖析我的代码
在某处放置
<pdb
断点,然后查看在哪个线程中哪个对象导致了违规(从而立即弄清楚它最初是如何在那里结束的)。使用错误消息以某种方式打印出对象(以及引用它们的父对象)
最佳答案
到目前为止,通过查看 pysqlite 的源代码,我可以说无法仅共享连接或游标对象(参见 pysqlite_check_thread
函数用法)。
pysqlite_check_thread
函数引发带有该消息的 ProgrammingError
异常
SQLite objects created in a thread can only be used in that same thread.
源代码中的某些函数捕获该异常并将其打印出来。
为了在您的源代码中找到在其他线程中调用连接方法的地方,我建议在连接对象上编写调用包装器,如下所示:
# script name: sqllitethread.py
import inspect
import sqlite3
import threading
import thread
from sqlite3 import ProgrammingError
class CallWrapper(object):
def __init__(self, obj):
self.obj = obj
self.main_thread_id = thread.get_ident()
def __getattr__(self, name):
if self.main_thread_id != thread.get_ident():
print "Thread %s requested `%s` attribute from %s" % (thread.get_ident(), name, self.obj)
for frame in inspect.getouterframes(inspect.currentframe())[1:]:
if frame[1].endswith('threading.py'):
continue
print "\t", "%s:%s" % (frame[1], frame[2]), frame[3], frame[4][0].strip()
return getattr(self.obj, name)
conn = CallWrapper(sqlite3.connect('example.db'))
c = conn.cursor()
def worker():
try:
conn.execute('.tables')
except ProgrammingError, e:
print e
t = threading.Thread(target=worker)
t.start()
t.join()
输出示例:
Thread 140390877370112 requested `execute` attribute from <sqlite3.Connection object at 0x7faf4e659858>
sqllitethread.py:30 worker conn.execute('.tables')
SQLite objects created in a thread can only be used in that same thread.The object was created in thread id 140390912665408 and this is thread id 140390877370112
关于python - 捕获导致 SQLite 多线程访问错误的对象,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/27884199/