Question

我正在尝试创建一个Web服务来将文件上传到服务器。我无法点击该服务并始终得到404错误，即找不到资源。我无法弄清楚我的代码中有什么问题。请帮忙。

下面是我的web.xml。我还有其他资源，如登录，注册和已注册的轨道，这些资源通过ajax请求工作得非常好，但是上传器web服务无效。

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">  <display-name>mixtri</display-name>
  <welcome-file-list>
    <welcome-file>index.html</welcome-file>
    <welcome-file>index.htm</welcome-file>
    <welcome-file>index.jsp</welcome-file>
    <welcome-file>default.html</welcome-file>
    <welcome-file>default.htm</welcome-file>
    <welcome-file>default.jsp</welcome-file>
  </welcome-file-list>
  <display-name>mixtri</display-name>
 <servlet>
    <servlet-name>Jersey REST Service</servlet-name>
    <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
     <!-- Register resources and providers under com.vogella.jersey.first package. -->
    <init-param>
        <param-name>jersey.config.server.provider.packages</param-name>
        <param-value>com.mixtri.tracks,com.mixtri.login,com.mixtri.signup,com.mixtri.uploader</param-value>
    </init-param>
    <init-param>
            <param-name>jersey.config.server.provider.classnames</param-name>
            <param-value>org.glassfish.jersey.media.multipart.MultiPartFeature</param-value>
     </init-param>
    <load-on-startup>1</load-on-startup>
  </servlet>
  <servlet-mapping>
    <servlet-name>Jersey REST Service</servlet-name>
    <url-pattern>/rest/*</url-pattern>
  </servlet-mapping>
</web-app>

以下是我的html表单的标记：

<form method="post" action="/mixtri/rest/upload"
                enctype="multipart/form-data">
                <table align="center" border="1" bordercolor="black" cellpadding="0"
                    cellspacing="0">
                    <tr>
                        <td>Select Zip File :</td>
                        <td><input type="file" name="uploadFile" size="100" /></td>
                    </tr>
                    <tr>
                        <td><input type="submit" value="Upload File" /></td>
                        <td><input type="reset" value="Reset" /></td>
                    </tr>
                </table>
            </form>

这是我的资源类：

package com.mixtri.uploader;

import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;

import javax.ws.rs.Consumes;
import javax.ws.rs.GET;
import javax.ws.rs.POST;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
import javax.ws.rs.core.Response;
import javax.ws.rs.core.Response.ResponseBuilder;

import org.glassfish.jersey.media.multipart.FormDataContentDisposition;
import org.glassfish.jersey.media.multipart.FormDataParam;

@Path("/rest")
public class FileServiceImpl implements IFileService {

    public static final String UPLOAD_FILE_SERVER = "D:/Demo/upload/";

    @POST
    @Path("/upload")
    @Consumes(MediaType.MULTIPART_FORM_DATA)
    public Response uploadZippedFile(
            @FormDataParam("uploadFile") InputStream fileInputStream,
            @FormDataParam("uploadFile") FormDataContentDisposition fileFormDataContentDisposition) {

        // local variables
        String fileName = null;
        String uploadFilePath = null;

        try {
            fileName = fileFormDataContentDisposition.getFileName();
            uploadFilePath = writeToFileServer(fileInputStream, fileName);
        }
        catch(IOException ioe){
            ioe.printStackTrace();
        }
        finally{
            // release resources, if any
        }
        return Response.ok("File uploaded successfully at " + uploadFilePath).build();
    }

    /**
     * write input stream to file server
     * @param inputStream
     * @param fileName
     * @throws IOException
     */
    private String writeToFileServer(InputStream inputStream, String fileName) throws IOException {

        OutputStream outputStream = null;
        String qualifiedUploadFilePath = UPLOAD_FILE_SERVER + fileName;

        try {
            outputStream = new FileOutputStream(new File(qualifiedUploadFilePath));
            int read = 0;
            byte[] bytes = new byte[1024];
            while ((read = inputStream.read(bytes)) != -1) {
                outputStream.write(bytes, 0, read);
            }
            outputStream.flush();
        }
        catch (IOException ioe) {
            ioe.printStackTrace();
        }
        finally{
            //release resource, if any
            outputStream.close();
        }
        return qualifiedUploadFilePath;
    }
}

任何指针都会有所帮助。

Answer 1

您可以将var result = users.GroupJoin(details, user => user.Id, detail => detail.Id, (user, detail) => new { user.Id, user.Name, user.Age, Height = detail.SingleOrDefault(x => x.Key == "Height").Value, Eyes = detail.SingleOrDefault(x => x.Key == "Eyes").Value, Hair = detail.SingleOrDefault(x => x.Key == "Hair").Value, });中的@Path("/rest")更改为FileServiceImpl。或者您可以将其更改为@Path("/")并将其从@Path("/upload")方法中删除。

您已将uploadZippedFile定义为泽西岛应用的根映射

/rest

所以最终你的路径需要

<servlet-mapping>
    <servlet-name>Jersey REST Service</servlet-name>
    <url-pattern>/rest/*</url-pattern>
</servlet-mapping>

如果您没有更改注释。

找不到资源Jersey 2.x.

1 个答案: