如何在没有Maven的情况下在Eclipse中创建src/main/java和src/test/java我的默认Web应用程序项目？我当前的文件夹结构是
TestApp
-JavaResources
- src
- com.mypackage
-WebContent
我无法将日志消息写入syslog.任何帮助都会很棒.这是我简单的log4j程序
import org.apache.log4j.Logger;
import java.io.*;
import java.sql.SQLException;
import java.util.*;
public class log4jExample
{
/* Get actual class name to be printed on */
static Logger log = Logger.getLogger(log4jExample.class.getName());
public static void main(String[] args) throws IOException,SQLException
{
log.error("Hello this is an error message");
log.info("Hello this is an info message");
log.fatal("Fatal error message");
}
}
我的syslog属性文件
# configure the root logger
log4j.rootLogger=INFO, SYSLOG
# configure Syslog facility LOCAL1 appender
log4j.appender.SYSLOG=org.apache.log4j.net.SyslogAppender
log4j.appender.SYSLOG.threshold=WARN
log4j.appender.SYSLOG.syslogHost=localhost
log4j.appender.SYSLOG.facility=LOCAL4
log4j.appender.SYSLOG.layout=org.apache.log4j.PatternLayout
log4j.appender.SYSLOG.layout.conversionPattern=[%p] %c:%L - %m%n
我可以在hibernate中的@Embeddable类中使用@Embedded吗？
示例:A是diffirent类中的元素集合.
@Embeddable
class A {
@Embedded
B b;
}
@Embeddable
class B {
@Embedded
C c;
}
@Embeddable
class C {
@Embedded
D D;
}
@Embeddable
class D {
}
这种类似的东西在休眠中是否有效？第三级嵌套.
有没有办法比较mysql查询中生成的范围列？
SELECT ue.bundle,ue.timestamp,b.id,bv.id as bundleVersionId,bv.start_date,bv.end_date, bv.type,ue.type from (
SELECT bundle,timestamp,tenant, case when Document_Id ='' then 'potrait'
WHEN Document_Id<>'' then 'persisted' end as type from uds_expanded ) ue
JOIN bundle b on b.name=ue.bundle join bundle_version bv on b.id=bv.bundle_id
WHERE ue.tenant='02306' and ue.timestamp >= bv.start_date and ue.timestamp <=bv.end_date and **ue.type=bv.type ;**
当我尝试比较类型时,我收到以下错误
Error Code: 1267. Illegal mix of collations (utf8_general_ci,COERCIBLE) and (latin1_swedish_ci,IMPLICIT) for operation '=' 0.000 sec
任何人都可以帮我创建rails中的下拉列表.我有一个带有角色字段的用户表,我想创建一个带有值管理器,调查员,角色主管的下拉列表.如何将该值提取到parms [:role].我是铁杆新手.
我有一个表与用户的has_many关联的案例.我试图通过在用户显示页面中呈现它来显示与用户相关的所有案例.我无法在这里渲染案例.任何帮助,将不胜感激.
我的用户显示页面代码:
<table >
<tr>
<td >
<h1>
Welcome <%= current_user.first_name %> you are <%= current_user.role %>
</h1>
</td>
</tr>
<tr>
<td>
<% render @cases %>
</td>
</tr>
</table>
在用户中显示控制器:
def show
@user = User.find(params[:id])
@cases = @user.cases
@title = @user.first_name
end
我在views/cases中有一个文件_case.html.erb.
我找不到课程错误.无法通过类加载器访问该类,并且从gradle运行sonarrunner时会出现错误提示.
这些类作为依赖项包含在gradle的编译时间中,并且gradle构建工作正常.如何将这些类包含在声纳中？
14:53:24.322 INFO - Load module settings
14:53:24.335 INFO - Base dir: /Users/srao1/Documents/GradleTest/lds_warehouse/lds_warehouse_base
14:53:24.335 INFO - Working dir: /Users/srao1/Documents/GradleTest/lds_warehouse/build/sonar/lds_warehouse_lds_warehouse_base
14:53:24.336 INFO - Source paths: src
14:53:24.336 INFO - Test paths: test
14:53:24.336 INFO - Binary dirs: build/classes/main
14:53:24.336 INFO - Source encoding: UTF-8, default locale: en_US
14:53:24.336 INFO - Index files
14:53:24.444 INFO - 50 files indexed
14:53:24.593 INFO - Quality profile for java: Sonar way
14:53:24.604 INFO - Sensor JavaSquidSensor
14:53:24.620 WARN - sonar.binaries and sonar.libraries are deprecated …我想在创建新用户(例如地址,引脚,电话号码等)期间跳过少数属性的验证.
但是,当用户尝试编辑它时,仍需要在模型中执行其他验证.我尝试使用:on =>:update但这对我没有帮助.有什么建议 ？
我的代码:
validates :address, :presence => true, :length => { :maximum => 50 }, :on => :update
validates :city, :presence => true, :length => { :maximum => 50 }, :on => :update
validates :state, :presence => true, :length => { :maximum => 50 }, :on => :update
validates :zip, :presence => true, :numericality => true, :on => :update, :length => { :is => 5 }