有没有一种pythonic的方法可以尝试最大次数的尝试？[重复]

我有一个python脚本正在查询共享Linux主机上的MySQL服务器。出于某种原因，对MySQL的查询通常会返回“服务器已消失”错误：

_mysql_exceptions.OperationalError: (2006, 'MySQL server has gone away')

如果此后立即再次尝试查询，通常会成功。因此，我想知道python中是否有一种明智的方法来尝试执行查询，如果失败，则可以重试固定次数的尝试。可能我想让它尝试5次再完全放弃。

这是我的代码类型：

conn = MySQLdb.connect(host, user, password, database)
cursor = conn.cursor()

try:
    cursor.execute(query)
    rows = cursor.fetchall()
    for row in rows:
        # do something with the data
except MySQLdb.Error, e:
    print "MySQL Error %d: %s" % (e.args[0], e.args[1])

显然，我可以通过在except子句中进行另一次尝试来做到这一点，但这非常丑陋，而且我觉得必须有一种不错的方法来实现这一目标。

怎么样：

conn = MySQLdb.connect(host, user, password, database)
cursor = conn.cursor()
attempts = 0

while attempts < 3:
    try:
        cursor.execute(query)
        rows = cursor.fetchall()
        for row in rows:
            # do something with the data
        break
    except MySQLdb.Error, e:
        attempts += 1
        print "MySQL Error %d: %s" % (e.args[0], e.args[1])

以Dana的答案为基础，您可能想作为装饰者来做：

def retry(howmany):
    def tryIt(func):
        def f():
            attempts = 0
            while attempts < howmany:
                try:
                    return func()
                except:
                    attempts += 1
        return f
    return tryIt

然后...

@retry(5)
def the_db_func():
    # [...]

使用该decorator模块的增强版本

import decorator, time

def retry(howmany, *exception_types, **kwargs):
    timeout = kwargs.get('timeout', 0.0) # seconds
    @decorator.decorator
    def tryIt(func, *fargs, **fkwargs):
        for _ in xrange(howmany):
            try: return func(*fargs, **fkwargs)
            except exception_types or Exception:
                if timeout is not None: time.sleep(timeout)
    return tryIt

然后...

@retry(5, MySQLdb.Error, timeout=0.5)
def the_db_func():
    # [...]

要安装的decorator模块：

$ easy_install decorator

更新：重试库有一个维护得更好的分支，称为tenacity，它支持更多功能，并且通常更灵活。

是的，有一个retrying库，它具有一个装饰器，该装饰器实现可以组合的几种重试逻辑：

一些例子：

@retry(stop_max_attempt_number=7)
def stop_after_7_attempts():
    print "Stopping after 7 attempts"

@retry(wait_fixed=2000)
def wait_2_s():
    print "Wait 2 second between retries"

@retry(wait_exponential_multiplier=1000, wait_exponential_max=10000)
def wait_exponential_1000():
    print "Wait 2^x * 1000 milliseconds between each retry,"
    print "up to 10 seconds, then 10 seconds afterwards"

conn = MySQLdb.connect(host, user, password, database)
cursor = conn.cursor()

for i in range(3):
    try:
        cursor.execute(query)
        rows = cursor.fetchall()
        for row in rows:
            # do something with the data
        break
    except MySQLdb.Error, e:
        print "MySQL Error %d: %s" % (e.args[0], e.args[1])

我会像这样重构它：

def callee(cursor):
    cursor.execute(query)
    rows = cursor.fetchall()
    for row in rows:
        # do something with the data

def caller(attempt_count=3, wait_interval=20):
    """:param wait_interval: In seconds."""
    conn = MySQLdb.connect(host, user, password, database)
    cursor = conn.cursor()
    for attempt_number in range(attempt_count):
        try:
            callee(cursor)
        except MySQLdb.Error, e:
            logging.warn("MySQL Error %d: %s", e.args[0], e.args[1])
            time.sleep(wait_interval)
        else:
            break

分解出callee功能似乎打破了功能，因此，它很容易看到业务逻辑，而在重试代码陷入困境。

像S.Lott一样，我喜欢一个标志来检查是否完成：

conn = MySQLdb.connect(host, user, password, database)
cursor = conn.cursor()

success = False
attempts = 0

while attempts < 3 and not success:
    try:
        cursor.execute(query)
        rows = cursor.fetchall()
        for row in rows:
            # do something with the data
        success = True 
    except MySQLdb.Error, e:
        print "MySQL Error %d: %s" % (e.args[0], e.args[1])
        attempts += 1

def successful_transaction(transaction):
    try:
        transaction()
        return True
    except SQL...:
        return False

succeeded = any(successful_transaction(transaction)
                for transaction in repeat(transaction, 3))

1.定义：

def try_three_times(express):
    att = 0
    while att < 3:
        try: return express()
        except: att += 1
    else: return u"FAILED"

2.用途

try_three_times(lambda: do_some_function_or_express())

我将其用于解析html上下文。

这是我的通用解决方案：

class TryTimes(object):
    ''' A context-managed coroutine that returns True until a number of tries have been reached. '''

    def __init__(self, times):
        ''' times: Number of retries before failing. '''
        self.times = times
        self.count = 0

    def __next__(self):
        ''' A generator expression that counts up to times. '''
        while self.count < self.times:
            self.count += 1
        yield False

    def __call__(self, *args, **kwargs):
        ''' This allows "o() calls for "o = TryTimes(3)". '''
        return self.__next__().next()

    def __enter__(self):
        ''' Context manager entry, bound to t in "with TryTimes(3) as t" '''
        return self

    def __exit__(self, exc_type, exc_val, exc_tb):
        ''' Context manager exit. '''
        return False # don't suppress exception

这允许如下代码：

with TryTimes(3) as t:
    while t():
        print "Your code to try several times"

也可能：

t = TryTimes(3)
while t():
    print "Your code to try several times"

我希望可以通过以更直观的方式处理异常来改善这一点。公开征求意见。

您可以使用for带有else子句的循环以达到最大效果：

conn = MySQLdb.connect(host, user, password, database)
cursor = conn.cursor()

for n in range(3):
    try:
        cursor.execute(query)
    except MySQLdb.Error, e:
        print "MySQL Error %d: %s" % (e.args[0], e.args[1])
    else:
        rows = cursor.fetchall()
        for row in rows:
            # do something with the data
        break
else:
    # All attempts failed, raise a real error or whatever

关键是查询成功后立即跳出循环。else仅当循环完成时不带时，才会触发该子句break。