is there a pythonic way to try something up to a maximum number of times?

Ben_11522 0
Viewed 69135

I have a python script which is querying a MySQL server on a shared linux host. For some reason, queries to MySQL often return a "server has gone away" error:

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

If you try the query again immediately afterwards, it usually succeeds. So, I'd like to know if there's a sensible way in python to try to execute a query, and if it fails, to try again, up to a fixed number of tries. Probably I'd want it to try 5 times before giving up altogether.

Here's the kind of code I have:

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])

Clearly I could do it by having another attempt in the except clause, but that's incredibly ugly, and I have a feeling there must be a decent way to achieve this.

exception-handling python
10 Answers

How about:

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])
Dana_78560

Building on Dana's answer, you might want to do this as a decorator:

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

Then...

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

Enhanced version that uses the decorator module

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

Then...

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

To install the decorator module:

$ easy_install decorator
dwc_573010 
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])
user_624210

I'd refactor it like so:

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

Factoring out the callee function seems to break up the functionality so that it's easy to see the business logic without getting bogged down in the retry code.

cdleary_35940

Like S.Lott, I like a flag to check if we're done:

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
user_495590

You can use a for loop with an else clause for maximum effect:

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

The key is to break out of the loop as soon as the query succeeds. The else clause will only be triggered if the loop completes without a break.

user_29887300
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