使用jersey上传Http 415

小编典典

使用jersey上传Http 415

jsp

我的RESTful文件上传代码：

@Path("/upload") 
@POST 
@Consumes("multipart/form-data") 
public String post(
    @FormDataParam("part") String s, 
    @FormDataParam("part") FormDataContentDisposition d) { 
    return s + ":" + d.getFileName(); 
}

当我尝试使用curl curl -X POST –form part=@file.txt url上传文件时

我收到HTTP 415不支持的媒体类型错误。怎么了？

阅读 495

2020-06-08

共1个答案

小编典典

在尝试了许多示例之后，请最终在 http://iambigd.blogspot.com/2011/06/java-upload-file-using-
jersey.html上找到真正有效的示例

@POST
@Path("/simpleupload")
@Consumes(MediaType.MULTIPART_FORM_DATA)
public void simpleUpload(
    //@Context UriInfo ui,
    @Context HttpServletRequest request
){
    String fileRepository = "D:\\";
    if (ServletFileUpload.isMultipartContent(request)) {
    FileItemFactory factory = new DiskFileItemFactory();
    ServletFileUpload upload = new ServletFileUpload(factory);
    List<FileItem> items = null;
    try {
        items = upload.parseRequest(request);
    } catch (FileUploadException e) {
        e.printStackTrace();
    }
    if (items != null) {
        Iterator<FileItem> iter = items.iterator();
        while (iter.hasNext()) {
            FileItem item = iter.next();
            if (!item.isFormField() && item.getSize() > 0) {
            System.out.println("File is found.");
            String fileName = processFileName(item.getName());
            try {
                String savePath = fileRepository + fileName;
                System.out.println("savePath:" + savePath);
                item.write(new File(savePath));
            } catch (Exception e) {
                e.printStackTrace();
            }
        }else{
            System.out.println("getFieldName:" + item.getFieldName());
            System.out.println(item.getString());
        }
     }
   }
}
}

（需要servlet-api.jar，（apache）commons-oi.jar和（apache）commons-fileupload.jar）

2020-06-08