小编Ana*_*ari的帖子

得到apache tomcat错误,访问被拒绝此文件localhost_access_log.2014-08-30.txt?

我在Windows 7操作系统上安装了Apache tomcate.我刚安装了apache并在eclipse中使用它.当我在服务器上运行任何简单的应用程序时,它会说404:找不到页面,在控制台中它将打印出这样的错误消息.

SEVERE: Failed to open access log file [C:\Program Files\Apache Software           Foundation\Tomcat 7.0\logs\localhost_access_log.2014-08-30.txt]
java.io.FileNotFoundException: C:\Program Files\Apache Software Foundation\Tomcat      7.0\logs\localhost_access_log.2014-08-30.txt (Access is denied)
at java.io.FileOutputStream.open(Native Method)
at java.io.FileOutputStream.<init>(Unknown Source)
at org.apache.catalina.valves.AccessLogValve.open(AccessLogValve.java:1117)
at org.apache.catalina.valves.AccessLogValve.startInternal(AccessLogValve.java:1224)
at org.apache.catalina.util.LifecycleBase.start(LifecycleBase.java:150)
at org.apache.catalina.core.StandardPipeline.startInternal(StandardPipeline.java:185)
at org.apache.catalina.util.LifecycleBase.start(LifecycleBase.java:150)
at org.apache.catalina.core.ContainerBase.startInternal(ContainerBase.java:1137)
at org.apache.catalina.core.StandardHost.startInternal(StandardHost.java:800)
at org.apache.catalina.util.LifecycleBase.start(LifecycleBase.java:150)
at org.apache.catalina.core.ContainerBase$StartChild.call(ContainerBase.java:1559)
at org.apache.catalina.core.ContainerBase$StartChild.call(ContainerBase.java:1549)
at java.util.concurrent.FutureTask$Sync.innerRun(Unknown Source)
at java.util.concurrent.FutureTask.run(Unknown Source)
at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
at java.lang.Thread.run(Unknown Source)
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我无法理解这是怎么回事?请帮我.

java apache tomcat

15
推荐指数
2
解决办法
3万
查看次数

如何在java中的String变量中获取异常消息?

当Java中捕获到任何异常时,我需要处理异常消息.

我正在研究数据库连接类.当我提供错误的详细信息,如用户名,密码,主机名,sid等,控件进入catch块并给出错误.我想在JSP端获取此错误消息并重定向到具有该错误消息的同一页面.但是,当我在Java中收到错误消息时,它总是需要一个空值.

我的代码示例在这里.

String errorMessage = null;
try{
// CODE Where Exception occure
}catch(SQLException se){
    errorMessage = se.getMessage();
}catch(Exception e){
    System. out.println("In Exception block.");
    errorMessage = e.getMessage();
}finally{
    System.out.println(errorMessage);
}
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它将转到Exception块,但errorMessage为null.

java

8
推荐指数
2
解决办法
5万
查看次数

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apache ×1

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