File upload 如何编写JAX-RS资源以使用上传的文件_File Upload_Jax Rs

File upload 如何编写JAX-RS资源以使用上传的文件

file-upload

File upload 如何编写JAX-RS资源以使用上传的文件,file-upload,jax-rs,File Upload,Jax Rs,这是我的班级：软件包rest 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.POST; import javax.ws.rs.Path; import javax.ws.

这是我的班级：

软件包rest

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.POST;
import javax.ws.rs.Path;
import javax.ws.rs.core.MediaType;
import javax.ws.rs.core.Response;
import com.sun.jersey.core.header.FormDataContentDisposition;
import com.sun.jersey.multipart.FormDataParam;

@Path("/file")
public class UploadFileService {

    @POST
    @Path("/upload")
    @Consumes(MediaType.MULTIPART_FORM_DATA)
    public Response uploadFile(
            @FormDataParam("file") InputStream uploadedInputStream,
            @FormDataParam("file") FormDataContentDisposition fileDetail) {

        String uploadedFileLocation = "d://uploaded/"
                + fileDetail.getFileName();

        // save it
        writeToFile(uploadedInputStream, uploadedFileLocation);

        String output = "File uploaded to : " + uploadedFileLocation;

        return Response.status(200).entity(output).build();

    }

    // save uploaded file to new location
    private void writeToFile(InputStream uploadedInputStream,
            String uploadedFileLocation) {

        try {
            OutputStream out = new FileOutputStream(new File(
                    uploadedFileLocation));
            int read = 0;
            byte[] bytes = new byte[1024];

            out = new FileOutputStream(new File(uploadedFileLocation));
            while ((read = uploadedInputStream.read(bytes)) != -1) {
                out.write(bytes, 0, read);
            }
            out.flush();
            out.close();
        } catch (IOException e) {

            e.printStackTrace();
        }

    }

}

这是我上传的Html文件：

<html>
<body>


    <h1>File Upload with Jersey</h1>

    <form action="rest/file/upload" method="post" enctype="multipart/form-data">

       <p>
        Select a file : <input type="file" name="file" size="45" />
       </p>

       <input type="submit" value="Upload It" />
    </form>

</body>
</html>

Web.xml文件：

<?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" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" 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>REST_WS</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>
</web-app>

当我尝试将文件上传到服务器时，没有任何操作执行我无法创建客户端直接在tomcat服务器中运行它html文件并尝试访问它请帮助我我们需要做什么更改我第一次写这个，所以请帮助。

您没有在web.xml中放入足够的内容，因此，容器不知道如何将请求路由到Jersey servlet

将以下内容添加到web.xml中：

您需要在所有要作为服务处理的url中使用/rest/*模式。

我们需要创建客户端或直接运行此项目？直接在tomcatStandardWrapper中运行此项目。一次性Exception即将出现，并且此异常也在上载映像后出现：为servlet UploadServlet分配异常您确定吗Jersey JAR在tomcat的lib目录或WEB-INF/lib目录中？

  <servlet>
    <servlet-name>UploadServlet</servlet-name>
    <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
    <init-param>
      <param-name>com.sun.jersey.config.property.packages</param-name>
      <param-value>pkg.rest</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
  </servlet>
  <servlet-mapping>
    <servlet-name>UploadServlet</servlet-name>
    <url-pattern>/rest/*</url-pattern>
  </servlet-mapping>