Java 使用PC remote读取JSP页面上的文件时出错
我有一个问题: 我有一个jsp客户端,它需要输入一个字符串来指示文件的路径:Java 使用PC remote读取JSP页面上的文件时出错,java,string,web-services,jsp,fileinputstream,Java,String,Web Services,Jsp,Fileinputstream,我有一个问题: 我有一个jsp客户端,它需要输入一个字符串来指示文件的路径: <%@page contentType="text/html" pageEncoding="UTF-8"%> <!DOCTYPE html> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <titl
<%@page contentType="text/html" pageEncoding="UTF-8"%>
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>JSP Page</title>
</head>
<body>
<form action="action.jsp" method="post">
Enter NameFile (Excel):<input type="text" name="filename"/><br/>
<input type="submit" value="Submit"/>
<input type="reset" value="Reset" />
</form>
</body>
</html>
JSP页面
输入名称文件(Excel):
现在,当我点击“提交”时,我看到了这个页面:
<%@page import="java.io.ByteArrayOutputStream"%>
<%@page import="java.io.FileInputStream"%>
<%@page import="java.io.File"%>
<%@page contentType="text/plain" pageEncoding="UTF-8"%>
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/plain; charset=UTF-8">
<title>JSP Page Action</title>
</head>
<body>
<%-- start web service invocation --%><hr/>
<%
String filename = request.getParameter("filename");
File file = new File (filename);
FileInputStream fis = new FileInputStream(file);
ByteArrayOutputStream bos = new ByteArrayOutputStream();
byte[] buf = new byte[1024];
for (int readNum; (readNum = fis.read(buf)) != -1;) {
bos.write(buf, 0, readNum); //no doubt here is 0
}
byte[] bi = bos.toByteArray();
try {
ws.MyWS_Service service = new ws.MyWS_Service();
ws.MyWS port = service.getMyWSPort();
// TODO initialize WS operation arguments here
byte[] inputFile = bi;
// TODO process result here
java.lang.String result = port.mywsmethod(inputFile);
out.println(result);
} catch (Exception ex) {
// TODO handle custom exceptions here
}
%>
<%-- end web service invocation --%><hr/>
</body>
</html>
JSP页面操作
现在,如果我插入
C:\prova.xls
在PC遥控器中,我遇到以下错误:
07-Oct-2014 08:55:10.194 SEVERE [http-apr-8080-exec-41] org.apache.catalina.core.StandardWrapperValve.invoke Servlet.service() for servlet [jsp] in context with path [/myWS_Client] threw exception [An exception occurred processing JSP page /action1.jsp at line 27
25:
26: File file = new File (filename);
27: FileInputStream fis = new FileInputStream(filename);
28: ByteArrayOutputStream bos = new ByteArrayOutputStream();
Stacktrace:] with root cause
java.io.FileNotFoundException: C:\prova.xls (Impossibile trovare il file specificato)
at java.io.FileInputStream.open(Native Method)
at java.io.FileInputStream.<init>(FileInputStream.java:146)
at java.io.FileInputStream.<init>(FileInputStream.java:101)
at org.apache.jsp.action1_005f1_jsp._jspService(action1_005f1_jsp.java:96)
at org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:70)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:725)
at org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:432)
at org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:391)
at org.apache.jasper.servlet.JspServlet.service(JspServlet.java:335)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:725)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:291)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
at org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:52)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:239)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:219)
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:106)
at org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:505)
at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:142)
at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:79)
at org.apache.catalina.valves.AbstractAccessLogValve.invoke(AbstractAccessLogValve.java:610)
at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:88)
at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:534)
at org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1081)
at org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:658)
at org.apache.coyote.http11.Http11AprProtocol$Http11ConnectionHandler.process(Http11AprProtocol.java:277)
at org.apache.tomcat.util.net.AprEndpoint$SocketProcessor.doRun(AprEndpoint.java:2381)
at org.apache.tomcat.util.net.AprEndpoint$SocketProcessor.run(AprEndpoint.java:2370)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1145)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:615)
at org.apache.tomcat.util.threads.TaskThread$WrappingRunnable.run(TaskThread.java:61)
at java.lang.Thread.run(Thread.java:744)
07-Oct-2014 08:55:10.194严重[http-apr-8080-exec-41]org.apache.catalina.core.StandardWrapperValve.invoke Servlet.service(),用于路径为[/myWS_Client]的上下文中的Servlet[jsp]抛出异常[在第27行处理jsp页/action1.jsp时发生异常]
25:
26:文件=新文件(文件名);
27:FileInputStream fis=新的FileInputStream(文件名);
28:ByteArrayOutputStream bos=新的ByteArrayOutputStream();
Stacktrace:]具有根本原因
java.io.FileNotFoundException:C:\prova.xls(不可能的trovare il文件规范)
在java.io.FileInputStream.open(本机方法)
位于java.io.FileInputStream。(FileInputStream.java:146)
位于java.io.FileInputStream。(FileInputStream.java:101)
在org.apache.jsp.action1\u 005f1\u jsp.\u jsp服务上(action1\u 005f1\u jsp.java:96)
位于org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:70)
位于javax.servlet.http.HttpServlet.service(HttpServlet.java:725)
位于org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:432)
位于org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:391)
位于org.apache.jasper.servlet.JspServlet.service(JspServlet.java:335)
位于javax.servlet.http.HttpServlet.service(HttpServlet.java:725)
位于org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:291)
位于org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
位于org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:52)
位于org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:239)
位于org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
位于org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:219)
位于org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:106)
位于org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:505)
位于org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:142)
位于org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:79)
位于org.apache.catalina.valves.AbstractAccessLogValve.invoke(AbstractAccessLogValve.java:610)
位于org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:88)
位于org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:534)
位于org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1081)
位于org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:658)
位于org.apache.coyote.http11.http11aprotocol$Http11ConnectionHandler.process(http11aprotocol.java:277)
位于org.apache.tomcat.util.net.aprendop$SocketProcessor.doRun(aprendop.java:2381)
位于org.apache.tomcat.util.net.aprendop$SocketProcessor.run(aprendop.java:2370)
位于java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1145)
位于java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:615)
位于org.apache.tomcat.util.threads.TaskThread$WrappingRunnable.run(TaskThread.java:61)
运行(Thread.java:744)
为什么我会犯这个错误?我怎么了?我试图用“/”、“\”、“\”和“/”来写这个路径,但我总是得到相同的错误。JSP是一个服务器端进程。服务器上可能不存在在HTML/Javascript中输入的文件名和路径 考虑使用文件上载将文件上载到JSP可以对其进行操作的服务器
也考虑使用servlet技术,在JSP中使用代码是不符合的。考虑一下,我想上传一个文件,把它转换成数组字节并发送到我的Web服务…我怎么做?我看到这个文件。。。但是我不想把文件上传到服务器上的某个目的地。。。我只想加载一个文件,转换并发送数组字节。文件与其单个字节集合之间的区别是什么?区别在于我无法将文件保存在服务器上,但是,我发送单个字节并在服务器上对其进行详细说明,而不保存它。我不想在服务器上保存任何内容根据我给你的链接,文件不需要保存在服务器上,只需直接使用inputstream
final Part filePart=request.getPart(“文件”)代码>filecontent=filePart.getInputStream()代码>