小编Abh*_*ngh的帖子

我有一个简单的"用户"集合,其中我现在只有2个文档.

{
    "_id": ObjectId("4ef8e1e41d41c87069000074"),
    "email_id": {
        "0": 109,
        "1": 101,
        "2": 64,
        "3": 97,

{
    "_id": ObjectId("4ef6d2641d41c83bdd000001"),
    "email_id": {
        "0": 109,
        "1": 97,
        "2": 105,
        "3": 108,

现在,如果我尝试在email_id字段上使用{unique:true}创建一个新索引,mongodb会向我抱怨"E11000重复键错误索引:db.users.$ email_id dup key:{:46}".即使在指定{dropDups:true}之后我也会得到相同的错误,但我不认为这是这种情况,因为两个文档都存储了不同的电子邮件ID.

我不确定这里发生了什么,任何指针都会非常感激.

编辑:文档的完整视图:

{
"_id": ObjectId("4ef8e1e41d41c87069000074"),
"email_id": {
 "0": 109,
 "1": 101,
 "2": 64,
 "3": 97,
 "4": 98,
 "5": 104,
 "6": 105,
 "7": 110,
 "8": 97,
 "9": 118,
 "10": 115,
 "11": 105,
 "12": 110,
 "13": 103,
 "14": 104,
 "15": 46,
 "16": 99,
 "17": 111,
 "18": 109
 }
}

和

{ …

我希望使用adb screencap没有-p标志的实用程序.我想象输出将以原始格式转储,但看起来不像.我尝试用Pillow(python)库打开原始图像文件导致:

$ adb pull /sdcard/screenshot.raw screenshot.raw
$ python
>>> from PIL import Image
>>> Image.open('screenshot.raw')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/Users/....../lib/python2.7/site-packages/PIL/Image.py", line 2025, in open
    raise IOError("cannot identify image file")
IOError: cannot identify image file

发现不是这样读取原始图像的正确方法,我甚至给了以下镜头:如何使用PIL读取原始图像？

>>> with open('screenshot.raw', 'rb') as f:
...     d = f.read()
... 
>>> from PIL import Image
>>> Image.frombuffer('RGB', len(d), d)
__main__:1: RuntimeWarning: the frombuffer defaults may change in a …

我有一个应用程序,其中每个websocket连接(在tornado打开回调内)创建一个zmq.SUB现有zmq.FORWARDER设备的套接字.想法是从zmq接收数据作为回调,然后可以通过websocket连接将其转发到前端客户端.

https://gist.github.com/abhinavsingh/6378134

ws.py

import zmq
from zmq.eventloop import ioloop
from zmq.eventloop.zmqstream import ZMQStream
ioloop.install()

from tornado.websocket import WebSocketHandler
from tornado.web import Application
from tornado.ioloop import IOLoop
ioloop = IOLoop.instance()

class ZMQPubSub(object):

    def __init__(self, callback):
        self.callback = callback

    def connect(self):
        self.context = zmq.Context()
        self.socket = self.context.socket(zmq.SUB)
        self.socket.connect('tcp://127.0.0.1:5560')
        self.stream = ZMQStream(self.socket)
        self.stream.on_recv(self.callback)

    def subscribe(self, channel_id):
        self.socket.setsockopt(zmq.SUBSCRIBE, channel_id)

class MyWebSocket(WebSocketHandler):

    def open(self):
        self.pubsub = ZMQPubSub(self.on_data)
        self.pubsub.connect()
        self.pubsub.subscribe("session_id")
        print 'ws opened'

    def on_message(self, message):
        print message

    def on_close(self):
        print 'ws …

我不久前在某处读过,可以配置外部jabber组件(XEP-0114)代表任何用户发送XMPP节.例如,假设我有一个组件绑定到(component.localhost),我希望它发送一个带有"from"属性设置为"user @ localhost"的消息节.

我试图用ejabberd实现这一目标.如果我不得不破解ejabberd src以使其工作(如果可能的话),那就不会感到惊讶了.