20

我正在开发一个应用程序,其工作流是通过使用 boto 在 SQS 中传递消息来管理的。

我的 SQS 队列正在逐渐增长,我无法检查它应该包含多少元素。

现在我有一个定期轮询队列的守护进程,并检查我是否有一组固定大小的元素。例如,考虑以下“队列”:

q = ["msg1_comp1", "msg2_comp1", "msg1_comp2", "msg3_comp1", "msg2_comp2"]

现在我想在某个时间点检查队列中是否有“msg1_comp1”、“msg2_comp1”和“msg3_comp1”,但我不知道队列的大小。

查看 API 后,您似乎只能获得 1 个元素,或者队列中的元素数量固定,但不是全部:

>>> rs = q.get_messages()
>>> len(rs)
1
>>> rs = q.get_messages(10)
>>> len(rs)
10

答案中提出的一个建议是在一个循环中获取例如 10 条消息,直到我什么也得不到,但是 SQS 中的消息有可见性超时,这意味着如果我从队列中轮询元素,它们不会被真正删除,它们只会在短时间内不可见。

有没有一种简单的方法来获取队列中的所有消息,而不知道有多少?

4

6 回答 6

23

把你的电话q.get_messages(n)放在while循环里面:

all_messages=[]
rs=q.get_messages(10)
while len(rs)>0:
    all_messages.extend(rs)
    rs=q.get_messages(10)

此外,转储也不支持超过 10 条消息

def dump(self, file_name, page_size=10, vtimeout=10, sep='\n'):
    """Utility function to dump the messages in a queue to a file
    NOTE: Page size must be < 10 else SQS errors"""
于 2012-04-16T20:06:24.347 回答
22

我一直在使用 AWS SQS 队列来提供即时通知,因此我需要实时处理所有消息。以下代码将帮助您有效地使(所有)消息出队并在删除时处理任何错误。

注意:要从队列中删除消息,您需要删除它们。我正在使用更新的 boto3 AWS python SDK、json 库和以下默认值:

import boto3
import json

region_name = 'us-east-1'
queue_name = 'example-queue-12345'
max_queue_messages = 10
message_bodies = []
aws_access_key_id = '<YOUR AWS ACCESS KEY ID>'
aws_secret_access_key = '<YOUR AWS SECRET ACCESS KEY>'
sqs = boto3.resource('sqs', region_name=region_name,
        aws_access_key_id=aws_access_key_id,
        aws_secret_access_key=aws_secret_access_key)
queue = sqs.get_queue_by_name(QueueName=queue_name)
while True:
    messages_to_delete = []
    for message in queue.receive_messages(
            MaxNumberOfMessages=max_queue_messages):
        # process message body
        body = json.loads(message.body)
        message_bodies.append(body)
        # add message to delete
        messages_to_delete.append({
            'Id': message.message_id,
            'ReceiptHandle': message.receipt_handle
        })

    # if you don't receive any notifications the
    # messages_to_delete list will be empty
    if len(messages_to_delete) == 0:
        break
    # delete messages to remove them from SQS queue
    # handle any errors
    else:
        delete_response = queue.delete_messages(
                Entries=messages_to_delete)
于 2015-10-15T18:36:27.380 回答
8

我的理解是,SQS 服务的分布式特性几乎使您的设计不可行。每次调用 get_messages 时,您都在与一组不同的服务器通信,这些服务器将包含您的一些但不是全部消息。因此,不可能“不时签入”来设置一组特定的消息是否准备好,然后就接受这些消息。

您需要做的是不断地轮询,在所有消息到达时获取它们,并将它们本地存储在您自己的数据结构中。每次成功获取后,您可以检查您的数据结构以查看是否收集了完整的消息集。

请记住,消息无序到达,并且某些消息被传递两次,因为删除必须传播到所有 SQS 服务器,但随后的获取请求有时会击败删除消息。

于 2012-04-21T23:11:19.517 回答
3

我在 cronjob 中执行这个

from django.core.mail import EmailMessage
from django.conf import settings
import boto3
import json

sqs = boto3.resource('sqs', aws_access_key_id=settings.AWS_ACCESS_KEY_ID,
         aws_secret_access_key=settings.AWS_SECRET_ACCESS_KEY,
         region_name=settings.AWS_REGION)

queue = sqs.get_queue_by_name(QueueName='email')
messages = queue.receive_messages(MaxNumberOfMessages=10, WaitTimeSeconds=1)

while len(messages) > 0:
    for message in messages:
        mail_body = json.loads(message.body)
        print("E-mail sent to: %s" % mail_body['to'])
        email = EmailMessage(mail_body['subject'], mail_body['message'], to=[mail_body['to']])
        email.send()
        message.delete()

    messages = queue.receive_messages(MaxNumberOfMessages=10, WaitTimeSeconds=1)
于 2017-06-21T15:02:40.893 回答
0

注意:这不是对问题的直接回答。 相反,它是对@TimothyLiu 答案的补充,假设最终用户使用的Boto不是包(又名 Boto2)Boto3此代码是他的回答delete_messages中提到的调用的“Boto-2 化”


Boto(2) 调用delete_message_batch(messages_to_delete)wheremessages_to_delete是一个 对象dict,其 key:value 对应于id: receipt_handlepair 返回

AttributeError:“dict”对象没有属性“id”。

似乎delete_message_batch需要一个Message类对象;如果您一次删除超过 10 条“消息”,则复制Boto 源delete_message_batch并允许它使用非Message对象(ala boto3 )也会失败。所以,我不得不使用以下解决方法。

这里打印代码

from __future__ import print_function
import sys
from itertools import islice

def eprint(*args, **kwargs):
    print(*args, file=sys.stderr, **kwargs)

@static_vars(counter=0)
def take(n, iterable, reset=False):
    "Return next n items of the iterable as same type"
    if reset: take.counter = 0
    take.counter += n
    bob = islice(iterable, take.counter-n, take.counter)
    if isinstance(iterable, dict): return dict(bob)
    elif isinstance(iterable, list): return list(bob)
    elif isinstance(iterable, tuple): return tuple(bob)
    elif isinstance(iterable, set): return set(bob)
    elif isinstance(iterable, file): return file(bob)
    else: return bob

def delete_message_batch2(cx, queue, messages): #returns a string reflecting level of success rather than throwing an exception or True/False
  """
  Deletes a list of messages from a queue in a single request.
  :param cx: A boto connection object.
  :param queue: The :class:`boto.sqs.queue.Queue` from which the messages will be deleted
  :param messages: List of any object or structure with id and receipt_handle attributes such as :class:`boto.sqs.message.Message` objects.
  """
  listof10s = []
  asSuc, asErr, acS, acE = "","",0,0
  res = []
  it = tuple(enumerate(messages))
  params = {}
  tenmsg = take(10,it,True)
  while len(tenmsg)>0:
    listof10s.append(tenmsg)
    tenmsg = take(10,it)
  while len(listof10s)>0:
    tenmsg = listof10s.pop()
    params.clear()
    for i, msg in tenmsg: #enumerate(tenmsg):
      prefix = 'DeleteMessageBatchRequestEntry'
      numb = (i%10)+1
      p_name = '%s.%i.Id' % (prefix, numb)
      params[p_name] = msg.get('id')
      p_name = '%s.%i.ReceiptHandle' % (prefix, numb)
      params[p_name] = msg.get('receipt_handle')
    try:
      go = cx.get_object('DeleteMessageBatch', params, BatchResults, queue.id, verb='POST')
      (sSuc,cS),(sErr,cE) = tup_result_messages(go)
      if cS:
        asSuc += ","+sSuc
        acS += cS
      if cE:
        asErr += ","+sErr
        acE += cE
    except cx.ResponseError:
      eprint("Error in batch delete for queue {}({})\nParams ({}) list: {} ".format(queue.name, queue.id, len(params), params))
    except:
      eprint("Error of unknown type in batch delete for queue {}({})\nParams ({}) list: {} ".format(queue.name, queue.id, len(params), params))
  return stringify_final_tup(asSuc, asErr, acS, acE, expect=len(messages)) #mdel #res

def stringify_final_tup(sSuc="", sErr="", cS=0, cE=0, expect=0):
  if sSuc == "": sSuc="None"
  if sErr == "": sErr="None"
  if cS == expect: sSuc="All"
  if cE == expect: sErr="All"
  return "Up to {} messages removed [{}]\t\tMessages remaining ({}) [{}]".format(cS,sSuc,cE,sErr)
于 2016-11-16T17:10:12.503 回答
-1

类似下面的代码应该可以解决问题。抱歉,它在 C# 中,但转换为 python 应该不难。字典用于清除重复项。

    public Dictionary<string, Message> GetAllMessages(int pollSeconds)
    {
        var msgs = new Dictionary<string, Message>();
        var end = DateTime.Now.AddSeconds(pollSeconds);

        while (DateTime.Now <= end)
        {
            var request = new ReceiveMessageRequest(Url);
            request.MaxNumberOfMessages = 10;

            var response = GetClient().ReceiveMessage(request);

            foreach (var msg in response.Messages)
            {
                if (!msgs.ContainsKey(msg.MessageId))
                {
                    msgs.Add(msg.MessageId, msg);
                }
            }
        }

        return msgs;
    }
于 2015-02-07T00:31:21.427 回答