I need to backup the current db while logged into odoo. I should be able to do it using a button, so that suppose I click on the button, it works the same way as odoo default backup in manage databases, but I should be able to do it from within while logged in. Is there any way to achieve this? I do know that this is possible from outside odoo using bash but thats not what I want. By using
我需要在登录odoo时备份当前数据库。 我应该可以使用按钮来完成它，所以假设我点击按钮，它与管理数据库中的odoo默认备份的方式一样，但我应该可以在登录时从内部执行。 有什么办法可以做到这一点？ 我知道使用bash可以从odoo之外实现，但那不是我想要的。 通过使用此模块，您可以定期备份数据库 https://www.odoo.com/apps/modules/7.0/crontab_config/（v7） 你也可以测试这个模块 https://www.odoo.com/apps/modu
I have the attached C++ wrapper example for python: The member function (method) is static with default argument. So I use BOOST_PYTHON_MEMBER_FUNCTION_OVERLOADS to define the overload function. There is no compilation error, however when I call the static member function I got the error as follows: import boostPythonTest boostPythonTest.C.method("string") ---------------------------------
我有附加的Python包装的C ++包装示例：成员函数（方法）是静态的，具有默认参数。 所以我使用BOOST_PYTHON_MEMBER_FUNCTION_OVERLOADS来定义重载函数。 没有编译错误，但是当我调用静态成员函数时出现如下错误： import boostPythonTest boostPythonTest.C.method("string") --------------------------------------------------------------------------- ArgumentError Traceback (most recent call last) <ipython-
How can I make a fifo between two python processes, that allow dropping of lines if the reader is not able to handle the input? If the reader tries to read or readline faster then the writer writes, it should block. If the reader cannot work as fast as the writer writes, the writer should not block. Lines should not be buffered (except one line at a time) and only the last line written shoul
如何在两个python进程之间创建fifo，如果读者无法处理输入，那么允许删除行？ 如果读者试图read或readline那么作家写快，应该可以挡住。 如果读者不能像写作者写作那样快，作者就不应该阻止。 行不应该被缓存（除了在一次一行），并只写了应该由读者在下次收到的最后一行readline尝试。 这是可能的一个命名的先进先出，或有任何其他简单的方法来实现这个？ 以下代码使用命名的FIFO来允许两个脚本之间的通信。 如果读
I am experimenting with using fifos for IPC routines in Python, and have the following code to create the fifo, then start two threads to write to it at varying times and read continually from it to print out whatever is put on the fifo. It works when the fifo is read using os.open(); however, I'm reading through O'Reilly's "Programming Python 4th Edition" where they clai
我正在尝试在Python中为IPC例程使用fifos，并且有以下代码来创建fifo，然后启动两个线程在不同的时间写入它，并不断读取它以打印出放在fifo上的任何内容。 它在使用os.open（）读取fifo时工作。 然而，我在阅读O'Reilly的“Programming Python 4th Edition”，他们声称fifo可以通过“消费者”过程作为文本文件对象打开。 在这里，将“consumer”线程切换为“consumer2”函数的目标是尝试将fifo作为文本对象读取，而不是使用os.o
I use standard tips for customizing interactive Python session: $ cat ~/.bashrc export PYTHONSTARTUP=~/.pystartup $ cat ~/.pystartup import os import sys import atexit import readline import rlcompleter historyPath = os.path.expanduser("~/.pyhistory") def save_history(historyPath=historyPath): import readline readline.write_history_file(historyPath) if os.path.exists(historyPath
我使用标准提示来定制交互式Python会话： $ cat ~/.bashrc export PYTHONSTARTUP=~/.pystartup $ cat ~/.pystartup import os import sys import atexit import readline import rlcompleter historyPath = os.path.expanduser("~/.pyhistory") def save_history(historyPath=historyPath): import readline readline.write_history_file(historyPath) if os.path.exists(historyPath): readline.read_hist
Something very strange is happening when I open FIFOs (named pipes) in Python for writing. Consider what happens when I try to open a FIFO for writing in a interactive interpreter: >>> fifo_write = open('fifo', 'w') The above line blocks until I open another interpreter and type the following: >>> fifo_read = open('fifo', 'r') >>> fifo.read() I don't understand
当我在Python中打开FIFO（命名管道）进行写入时，发生了一些非常奇怪的事情。 考虑当我尝试打开一个用于在交互式解释器中编写的FIFO时会发生什么情况： >>> fifo_write = open('fifo', 'w') 上述行阻止，直到我打开另一个解释器并键入以下内容： >>> fifo_read = open('fifo', 'r') >>> fifo.read() 我不明白为什么我必须等待管道打开才能阅读，但让我们跳过这一点。 上述代码将会阻塞，直到有
Django's builtin capability of emailing admins upon errors (see https://docs.djangoproject.com/en/dev/howto/error-reporting/) is quite handy. However, these traceback emails include a full dump of environment variables. And as advised in the django docs & elsewhere (eg https://docs.djangoproject.com/en/dev/howto/deployment/checklist/) I've moved some secrets/keys/passwords into en
Django的内置功能可以在发生错误时向管理员发送电子邮件（请参阅https://docs.djangoproject.com/en/dev/howto/error-reporting/），非常方便。 但是，这些追溯电子邮件包含完整的环境变量转储。 正如django文档和其他地方（例如https://docs.djangoproject.com/en/dev/howto/deployment/checklist/）所建议的那样，我已经将一些秘密/密钥/密码移入环境变量中，作为一种简单的方法让它们远离代码库并在部署中改变它们。 不
I have a Celery Task-Manager to crunch some numbers for company analytics. The Task-Manager and workers are hosted on an Amazon EC2 Linux Server. I need to set up the system such if we send too many tasks to celery Amazon automatically sets up a new EC2 instance to run more workers and balances the load across these workers. The services that I'm aware exist are the Amazon Autoscale and
我有一个Celery任务管理器来为公司分析找出一些数字。 任务管理器和工作人员托管在Amazon EC2 Linux服务器上。 如果我们向芹菜发送太多任务，我需要建立系统。亚马逊会自动设置一个新的EC2实例来运行更多的员工，并平衡这些员工的负载。 我知道的服务存在的是Amazon Autoscale和Amazon负载平衡服务，看起来正是我想要使用的服务，但我不确定配置Celery的最佳方式是什么。 我认为我应该有一个收集所有任务的芹菜“主人”和
I am using heapq to get nlargest elements from list of lists. The program I wrote is below. import csv import heapq f = open("E:/output.csv","r") read = csv.reader(f) allrows = [row for row in read] for i in xrange(0,2): print allrows[i] allrows.sort(key=lambda x: x[2]) #this is working properly it=heapq.nlargest(20,enumerate(allrows),key=lambda x:x[2]) #error I just want the top 20 el
我正在使用heapq从列表中获取最大的元素。 我写的程序如下。 import csv import heapq f = open("E:/output.csv","r") read = csv.reader(f) allrows = [row for row in read] for i in xrange(0,2): print allrows[i] allrows.sort(key=lambda x: x[2]) #this is working properly it=heapq.nlargest(20,enumerate(allrows),key=lambda x:x[2]) #error 我只想要前20名元素。 所以，而不是排序我想用堆。 我得到的
Here is an idea for a dict subclass that can mutate keys. This is a simple self contained example that's just like a dict but is case insensitive for str keys. from functools import wraps def key_fix_decorator(f): @wraps(f) def wrapped(self, *args, **kwargs): if args and isinstance(args[0], str): args = (args[0].lower(),) + args[1:] return f(self, *args,
这是一个可以改变密钥的字典子类的想法。 这是一个简单的自包含的例子，就像dict但对str键不区分大小写。 from functools import wraps def key_fix_decorator(f): @wraps(f) def wrapped(self, *args, **kwargs): if args and isinstance(args[0], str): args = (args[0].lower(),) + args[1:] return f(self, *args, **kwargs) return wrapped class LowerDict(dict): pass f