Idea使用Tomcat时Servlet找不到xml properties文件

asdfagdsag 2018-03-04 07:17:51

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
Properties properties = new Properties();
properties.load(new FileInputStream("a.properties"));
}

如这段代码报错为:

java.io.FileNotFoundException: a.properties (系统找不到指定的文件。)
at java.io.FileInputStream.open0(Native Method)
at java.io.FileInputStream.open(FileInputStream.java:195)
at java.io.FileInputStream.<init>(FileInputStream.java:138)
at java.io.FileInputStream.<init>(FileInputStream.java:93)
at Servlet1.doPost(Servlet1.java:16)
at Servlet1.doGet(Servlet1.java:23)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:634)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:741)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:231)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:166)
at org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:53)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:193)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:166)
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:199)
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:96)
at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:137)
at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:81)
at org.apache.catalina.valves.AbstractAccessLogValve.invoke(AbstractAccessLogValve.java:651)
at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:87)
at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:342)
at org.apache.coyote.http11.Http11Processor.service(Http11Processor.java:409)
at org.apache.coyote.AbstractProcessorLight.process(AbstractProcessorLight.java:66)
at org.apache.coyote.AbstractProtocol$ConnectionHandler.process(AbstractProtocol.java:754)
at org.apache.tomcat.util.net.NioEndpoint$SocketProcessor.doRun(NioEndpoint.java:1376)
at org.apache.tomcat.util.net.SocketProcessorBase.run(SocketProcessorBase.java:49)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1149)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:624)
at org.apache.tomcat.util.threads.TaskThread$WrappingRunnable.run(TaskThread.java:61)
at java.lang.Thread.run(Thread.java:748)

a.properties此时无论放在resource目录下还是src根目录下都一样
------------------------------------------------------------------------------------------------------------------------------------------
然后,
System.getProperty("user.dir")

这段代码输出的路径为Tomcat的bin目录?????????
我只是一年没敲代码,发生了什么??????????
没有使用Maven
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WarsFeng 2018-03-05
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引用 1 楼 qq_34758081 的回复:
你试试把路径写全
没用只有丢在tomcat的bin目录下有用。。。。
qq_34758081 2018-03-05
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你试试把路径写全

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