在rails项目上配置日志功能的最佳方法是什么?我正在寻找Rails可用的Log4J之类的东西.我发现log4r并且它在Logger类中构建了冲突并且还尝试了"Logging"gem并且它在配置为审计记录器时存在一些问题.因为我是这个主题的初学者,请让我知道你对这个主题的建议.
我在logging.rb中使用了以下代码块并包含在environment.rb中但是我收到了'returns'关键字的错误,因为它在rails 2.8上已被弃用
# Logging
require File.join(File.dirname(__FILE__), 'logging')
Rails::Initializer.run do |config|
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require 'logging'
# Logging.init is required to avoid
# unknown level was given 'info' (ArgumentError)
# or
# uninitialized constant Logging::MAX_LEVEL_LENGTH (NameError)
# when an Appender or Layout is created BEFORE any Logger is instantiated:
Logging.init :debug, :info, :warn, :error, :fatal
layout = Logging::Layouts::Pattern.new :pattern => "[%d] [%-5l] %m\n"
# Default logfile, history kept for 10 days
default_appender = Logging::Appenders::RollingFile.new 'default', \
:filename => 'log/default.log', :age …
Run Code Online (Sandbox Code Playgroud) 我有一个ZF2项目,具有以下配置.它使用的是Doctrine ORM和SlmQueue.由于SlmQueue不支持我们的命名约定和oracle数据库,因此我们定制了SlmQueueDoctrine.
我怀疑我的工作不是使用ClearObjectManagerStrategy,并且在执行单个作业之前不清除ObjectManager.
它在队列启动后不反映数据库修改.但是如果我杀死队列守护进程并重新开始,它会选择新的值.
如何在执行单个作业之前实现ClearObjectManagerStrategy并清除ObjectManager?
我试过很多没有运气的人.
composer.json
{
"repositories": [
{
"url": "https://github.com/pradeep-sanjaya/doctrine-extensions.git",
"type": "git"
}
],
"require": {
"php": ">=5.3.3",
"zendframework/zendframework": "2.3.3",
"doctrine/doctrine-orm-module": "0.7.*",
"pradeep-sanjaya/doctrine-extensions": "dev-master",
"spoonx/sxmail": "1.4.*",
"slm/locale": "dev-master",
"imagine/Imagine": "0.6.*",
"tecnick.com/tcpdf": "dev-master",
"slm/queue": "0.4.*",
"slm/queue-doctrine": "0.4.*"
}
}
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配置/自动加载/ slm_queue.local.php
<?php
return array(
'slm_queue' => array(
'queue_manager' => array(
'factories' => array(
'doctrineQueue' => 'SlmQueueDoctrine\Factory\DoctrineQueueFactory'
),
),
'job_manager' => array(
'factories' => array(
'Report\Job\Rank' => 'Report\Job\RankFactory',
),
'shared' => array(
'Report\Job\Rank' => false
),
),
'queues' => …
Run Code Online (Sandbox Code Playgroud) 目前我正在使用:
$query = SELECT * FROM `table` WHERE `id`='column';
$mydata = mysql_query($query);
$element = mysql_result($mydata, 0, "Element");
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是否有更快,更好的方法来做同样的事情?