这是我的代码,我在shelve中更新记录
def updateRecord(db, form):
if not 'key' in form:
fields = dict.fromkeys(fieldnames, '?')
fields['key'] = 'Missing key input'
else:
key = form['key'].value
if key in db:
record = db[key]
else:
from person import Person
record = Person(name='?',age='?')
for field in fieldnames:
setattr(record, field, eval(form[field].value))
db[key] = record
fields = record.__dict__
fields['key'] = key
return fields
当我尝试从搁置中检索值时,我收到此错误
>>> import shelve
>>> db = shelve.open('class-shelve')
>>> db['sue'].name
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/python2.7/shelve.py", line 121, in __getitem__
f = StringIO(self.dict[key])
File "/usr/lib/python2.7/bsddb/__init__.py", line 270, in __getitem__
return _DeadlockWrap(lambda: self.db[key]) # self.db[key]
File "/usr/lib/python2.7/bsddb/dbutils.py", line 68, in DeadlockWrap
return function(*_args, **_kwargs)
File "/usr/lib/python2.7/bsddb/__init__.py", line 270, in <lambda>
return _DeadlockWrap(lambda: self.db[key]) # self.db[key]
KeyError: 'sue'
有什么见解?
答案 0 :(得分:1)
假设在第一个片段中,db
变量是一个“架子”对象,那么,虽然该行......
db[key] = record
...会将新的键/值对添加到'shelf',它不一定将内容刷新到磁盘,因此共享相同的'shelf文件'的其他进程将无法使用它。< / p>
您可以通过添加行...
强制将“shelf文件”写入磁盘db.sync()
...添加新的键/值对后,但当“架子文件”变大时,它可能会非常慢,所以您可能不想太频繁地调用它。