使用JAX-RS的FileUpload [英] FileUpload with JAX-RS
本文介绍了使用JAX-RS的FileUpload的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我尝试从JavaScript客户端上传文件到JAX-RS Java服务器。
我在服务器上使用以下REST上传功能：
@POST
@Produces（'application / json'）
UploadDto上传（
@Context HttpServletRequest请求，
@QueryParam（cookie）字符串cookie）{
$ b $ def contentType
字节[] fileBytes
log.debugupload - cookie ：+ cookie
尝试{
if（请求instanceof MultipartHttpServletRequest）{
log.debugrequest instanceof MultipartHttpServletRequest
MultipartHttpServletRequest myrequest = request
CommonsMultipartFile file =（CommonsMultipartFile）myrequest.getFile（'file'）
fileBytes = file.bytes
contentType = file.contentType
log.debug>>>> ;>以字节为单位上传文件的大小：+ file.size
}
（请求instanceof SecurityContextHolderAwareRequestWrapper）{
log.debugrequest instanceof SecurityContextHolderAwareRequestWrapper
SecurityContextHolderAwareRequestWrapper myrequest =请求
//获取上传文件的inputStream
InputStream inputStream = myrequest.inputStream
fileBytes = IOUtils.toByteArray（inputStream）;
contentType = myrequest.getHeader（Content-Type）
log.debug>>>>>>>上传文件大小的字节：+ fileBytes.size（）
else {
log.error请求不是MultipartHttpServletRequest或SecurityContextHolderAwareRequestWrapper
printlnrequest：+ request.class
}
}
catch（IOException e）{
log.error（upload（）无法保存文件错误：，e）
}
}
code> var str2ab_blobreader = function（str，callback）{
var blob;
BlobBuilder = window.MozBlobBuilder || window.WebKitBlobBuilder
|| window.BlobBuilder;
if（typeof（BlobBuilder）！=='undefined'）{
var bb = new BlobBuilder（）;
bb.append（str）;
blob = bb.getBlob（）;
} else {
blob = new Blob（[str]）;
}
var f = new FileReader（）;
f.onload = function（e）{
callback（e.target.result）
}
f.readAsArrayBuffer（blob）;
}
var fileName =fileName.jpg;
var contentType =image / jpeg;
if（file.type.toString（）。toLowerCase（）。indexOf（png）> -1）{
fileName =fileName.png;
contentType =image / png;
}
var xhrNativeObject = new XMLHttpRequest（）;
var urlParams =？test = 123;
xhrNativeObject.open（post，url + urlParams，true）;
xhrNativeObject.setRequestHeader（Content-Type，contentType）;
xhrNativeObject.onload = function（event）{
var targetResponse = event.currentTarget; （（targetResponse.readyState == 4）
&&（targetResponse.status == 200））{
var obj = JSON.parse（targetResponse.responseText）;
console.log（obj.uploadImageId）;
} else {
console.log（fail）;
var buffer = str2ab_blobreader（file，function（buf）{
xhrNativeObject.send（buf）;
}）;
当我在我的Grails控制器中使用代码时，它运行良好，但是当我在REST资源中使用它时我总是得到：请求不是MultipartHttpServletRequest或SecurityContextHolderAwareRequestWrapper
日志输出是
request：com.sun.proxy。$ Proxy58
从JavaScript发送文件blob我使用> XMLHttpRequest ，其中包含正文中的blob和一些查询参数。
我该如何制作JAX-RS文件上传工作？我如何收到一些额外的查询参数与我的POST请求？
解决方案
在服务器端，您可以使用类似这个
@POST
@Path（/ fileupload）//您的路径或URL调用此服务
@Consumes（MediaType.MULTIPART_FORM_DATA）
public response uploadFile（
@DefaultValue（true）@FormDataParam（enabled）boolean enabled，
@FormDataParam（file） InputStream uploadedInputStream，
@FormDataParam（file）FormDataContentDisposition fileDetail）{
//您要存储文件的本地磁盘路径
String uploadedFileLocation =D：// uploadedFiles / + fileDetail.getFileName（）;
System.out.println（uploadedFileLocation）;
//保存
文件objFile = new File（uploadedFileLocation）;
if（objFile.exists（））
{
objFile.delete（）;
}
saveToFile（uploadedInputStream，uploadedFileLocation）;
字符串输出=通过基于Jersey的RESTFul Webservice上传的文件：+ uploadedFileLocation;
return Response.status（200）.entity（output）.build（）;
$ b private void saveToFile（InputStream uploadedInputStream，
String uploadedFileLocation）{
try {
OutputStream out = null;
int read = 0;
byte [] bytes = new byte [1024];
out = new FileOutputStream（new File（uploadedFileLocation））; （（read = uploadedInputStream.read（bytes））！= -1）{
out.write（bytes，0，read）;
while
}
out.flush（）;
out.close（）;
} catch（IOException e）{
e.printStackTrace（）;
}
}
这可以用客户端代码在Java中与
public class TryFile {
public static void main（String [] ar）
抛出HttpException，IOException，URISyntaxException {
TryFile t = new TryFile（）;
t.method（）;
}
public void method（）抛出HttpException，IOException，URISyntaxException {
String url =http：// localhost：8080 /...../ fileupload; //你的服务URL
String fileName =; //要上传的文件名
HttpClient httpclient = new DefaultHttpClient（）;
HttpPost httppost = new HttpPost（url）;
FileBody fileContent = new FiSystem.out.println（hello）;
StringBody comment = new StringBody（Filename：+ fileName）;
MultipartEntity reqEntity = new MultipartEntity（）;
reqEntity.addPart（file，fileContent）;
httppost.setEntity（reqEntity）;
HttpResponse response = httpclient.execute（httppost）;
HttpEntity resEntity = response.getEntity（）;
$ b $ p
$ b 使用HTML，你可以简单地检查这段代码
< html>
< body>
< h1>使用RESTFul WebService上传文件< / h1>
< form action =< Your service URL（htp：// localhost：8080 /.../ fileupload）method =postenctype =multipart / form-data>
< p>
选择一个文件：< input type =filename =file/>
< / p>
< input type =submitvalue =上传/>
< / form>
QueryParam，Check @QueryParam或头部参数使用@HeaderParam
试试这个，希望这可以帮助你问题。
I try to do file upload from a JavaScript client to a JAX-RS Java server.
I use the following REST upload function on my server:
@POST
@Produces('application/json')
UploadDto upload(
@Context HttpServletRequest request,
@QueryParam("cookie") String cookie) {
def contentType
byte [] fileBytes
log.debug "upload - cookie: "+cookie
try{
if (request instanceof MultipartHttpServletRequest) {
log.debug "request instanceof MultipartHttpServletRequest"
MultipartHttpServletRequest myrequest = request
CommonsMultipartFile file = (CommonsMultipartFile) myrequest.getFile('file')
fileBytes = file.bytes
contentType = file.contentType
log.debug ">>>>> upload size of the file in byte: "+ file.size
}
else if (request instanceof SecurityContextHolderAwareRequestWrapper) {
log.debug "request instanceof SecurityContextHolderAwareRequestWrapper"
SecurityContextHolderAwareRequestWrapper myrequest = request
//get uploaded file's inputStream
InputStream inputStream = myrequest.inputStream
fileBytes = IOUtils.toByteArray(inputStream);
contentType = myrequest.getHeader("Content-Type")
log.debug ">>>>> upload size of the file in byte: "+ fileBytes.size()
}
else {
log.error "request is not a MultipartHttpServletRequest or SecurityContextHolderAwareRequestWrapper"
println "request: "+request.class
}
}
catch (IOException e) {
log.error("upload() failed to save file error: ", e)
}
}
On the client side I send the file as follows:
var str2ab_blobreader = function(str, callback) {
var blob;
BlobBuilder = window.MozBlobBuilder || window.WebKitBlobBuilder
|| window.BlobBuilder;
if (typeof (BlobBuilder) !== 'undefined') {
var bb = new BlobBuilder();
bb.append(str);
blob = bb.getBlob();
} else {
blob = new Blob([ str ]);
}
var f = new FileReader();
f.onload = function(e) {
callback(e.target.result)
}
f.readAsArrayBuffer(blob);
}
var fileName = "fileName.jpg";
var contentType = "image/jpeg";
if (file.type.toString().toLowerCase().indexOf("png") > -1) {
fileName = "fileName.png";
contentType = "image/png";
}
var xhrNativeObject = new XMLHttpRequest();
var urlParams = ?test=123;
xhrNativeObject.open("post", url + urlParams, true);
xhrNativeObject.setRequestHeader("Content-Type", contentType);
xhrNativeObject.onload = function(event) {
var targetResponse = event.currentTarget;
if ((targetResponse.readyState == 4)
&& (targetResponse.status == 200)) {
var obj = JSON.parse(targetResponse.responseText);
console.log(obj.uploadImageId);
} else {
console.log("fail");
}
}
var buffer = str2ab_blobreader(file, function(buf) {
xhrNativeObject.send(buf);
});
When I use the code in my Grails Controller it worked well but when I use it in a REST Resource I always get: request is not a MultipartHttpServletRequest or SecurityContextHolderAwareRequestWrapper
The log output is
request: com.sun.proxy.$Proxy58
The send a file blob from JavaScript I use XMLHttpRequest which contains the blob in the body and some query parameters.
How can I make JAX-RS file upload working? How do I receive some additional query params with my POST request?
解决方案 On Server Side you can use something like this
@POST
@Path("/fileupload") //Your Path or URL to call this service
@Consumes(MediaType.MULTIPART_FORM_DATA)
public Response uploadFile(
@DefaultValue("true") @FormDataParam("enabled") boolean enabled,
@FormDataParam("file") InputStream uploadedInputStream,
@FormDataParam("file") FormDataContentDisposition fileDetail) {
//Your local disk path where you want to store the file
String uploadedFileLocation = "D://uploadedFiles/" + fileDetail.getFileName();
System.out.println(uploadedFileLocation);
// save it
File objFile=new File(uploadedFileLocation);
if(objFile.exists())
{
objFile.delete();
}
saveToFile(uploadedInputStream, uploadedFileLocation);
String output = "File uploaded via Jersey based RESTFul Webservice to: " + uploadedFileLocation;
return Response.status(200).entity(output).build();
}
private void saveToFile(InputStream uploadedInputStream,
String uploadedFileLocation) {
try {
OutputStream out = null;
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();
}
}
Again this can be checked with the client code in java with
public class TryFile {
public static void main(String[] ar)
throws HttpException, IOException, URISyntaxException {
TryFile t = new TryFile();
t.method();
}
public void method() throws HttpException, IOException, URISyntaxException {
String url = "http://localhost:8080/...../fileupload"; //Your service URL
String fileName = ""; //file name to be uploaded
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(url);
FileBody fileContent = new FiSystem.out.println("hello");
StringBody comment = new StringBody("Filename: " + fileName);
MultipartEntity reqEntity = new MultipartEntity();
reqEntity.addPart("file", fileContent);
httppost.setEntity(reqEntity);
HttpResponse response = httpclient.execute(httppost);
HttpEntity resEntity = response.getEntity();
}
}
With HTML, you can simply check with this code
<html>
<body>
<h1>Upload File with RESTFul WebService</h1>
<form action="<Your service URL (htp://localhost:8080/.../fileupload)" method="post" enctype="multipart/form-data">
<p>
Choose a file : <input type="file" name="file" />
</p>
<input type="submit" value="Upload" />
</form>
To get QueryParam, Check @QueryParam or for header param use @HeaderParam
Try this, hope this helps you with your problem.
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