Mixing
upload a file via RestEasy via MultipartOutput
I'm trying to programmatically upload a file to an enpoint via RestEasyClient.
File file = new File("/Users/michele/path/file.txt");
MultipartOutput multipartOutput = new MultipartOutput();
multipartOutput.addPart(file, MediaType.APPLICATION_OCTET_STREAM_TYPE, "file.txt");
Entity<MultipartOutput> entity = Entity.entity(multipartOutput, MediaType.MULTIPART_FORM_DATA_TYPE);
//client is an instance of org.jboss.resteasy.client.jaxrs.ResteasyClient
client
.target("http://localhost:8080/endpoint")
.request()
.post(entity);
The problem is that the backend does not "find" the file that I uploaded
backend code
DiskFileItemFactory factory = new DiskFileItemFactory();
ServletFileUpload fileUpload = new ServletFileUpload(factory);
List<FileItem> items = fileUpload.parseRequest(httpReq);
items is always empt.
Using MultipartFormDataOutput::addFormData, as described in many articles, works but does not fit my use case.
Also using apache.http.client.HttpClient works, but I prefer to avoid adding dependencies to my client.
Any ideas?
java resteasy
asked Jan 2 at 11:11
Michele Da RosMichele Da Ros
418215
add a comment |
I'm trying to programmatically upload a file to an enpoint via RestEasyClient.
File file = new File("/Users/michele/path/file.txt");
MultipartOutput multipartOutput = new MultipartOutput();
multipartOutput.addPart(file, MediaType.APPLICATION_OCTET_STREAM_TYPE, "file.txt");
Entity<MultipartOutput> entity = Entity.entity(multipartOutput, MediaType.MULTIPART_FORM_DATA_TYPE);
//client is an instance of org.jboss.resteasy.client.jaxrs.ResteasyClient
client
.target("http://localhost:8080/endpoint")
.request()
.post(entity);
The problem is that the backend does not "find" the file that I uploaded
backend code
DiskFileItemFactory factory = new DiskFileItemFactory();
ServletFileUpload fileUpload = new ServletFileUpload(factory);
List<FileItem> items = fileUpload.parseRequest(httpReq);
items is always empt.
Using MultipartFormDataOutput::addFormData, as described in many articles, works but does not fit my use case.
Also using apache.http.client.HttpClient works, but I prefer to avoid adding dependencies to my client.
Any ideas?
java resteasy
asked Jan 2 at 11:11
Michele Da RosMichele Da Ros
418215
add a comment |
I'm trying to programmatically upload a file to an enpoint via RestEasyClient.
File file = new File("/Users/michele/path/file.txt");
MultipartOutput multipartOutput = new MultipartOutput();
multipartOutput.addPart(file, MediaType.APPLICATION_OCTET_STREAM_TYPE, "file.txt");
Entity<MultipartOutput> entity = Entity.entity(multipartOutput, MediaType.MULTIPART_FORM_DATA_TYPE);
//client is an instance of org.jboss.resteasy.client.jaxrs.ResteasyClient
client
.target("http://localhost:8080/endpoint")
.request()
.post(entity);
The problem is that the backend does not "find" the file that I uploaded
backend code
DiskFileItemFactory factory = new DiskFileItemFactory();
ServletFileUpload fileUpload = new ServletFileUpload(factory);
List<FileItem> items = fileUpload.parseRequest(httpReq);
items is always empt.
Using MultipartFormDataOutput::addFormData, as described in many articles, works but does not fit my use case.
Also using apache.http.client.HttpClient works, but I prefer to avoid adding dependencies to my client.
Any ideas?
java resteasy
asked Jan 2 at 11:11
Michele Da RosMichele Da Ros
418215
I'm trying to programmatically upload a file to an enpoint via RestEasyClient.
File file = new File("/Users/michele/path/file.txt");
MultipartOutput multipartOutput = new MultipartOutput();
multipartOutput.addPart(file, MediaType.APPLICATION_OCTET_STREAM_TYPE, "file.txt");
Entity<MultipartOutput> entity = Entity.entity(multipartOutput, MediaType.MULTIPART_FORM_DATA_TYPE);
//client is an instance of org.jboss.resteasy.client.jaxrs.ResteasyClient
client
.target("http://localhost:8080/endpoint")
.request()
.post(entity);
The problem is that the backend does not "find" the file that I uploaded
backend code
DiskFileItemFactory factory = new DiskFileItemFactory();
ServletFileUpload fileUpload = new ServletFileUpload(factory);
List<FileItem> items = fileUpload.parseRequest(httpReq);
items is always empt.
Using MultipartFormDataOutput::addFormData, as described in many articles, works but does not fit my use case.
Also using apache.http.client.HttpClient works, but I prefer to avoid adding dependencies to my client.
Any ideas?
java resteasy
java resteasy
asked Jan 2 at 11:11
Michele Da RosMichele Da Ros
418215
asked Jan 2 at 11:11
Michele Da RosMichele Da Ros
418215
asked Jan 2 at 11:11
Michele Da RosMichele Da Ros
418215
asked Jan 2 at 11:11
Michele Da RosMichele Da Ros
418215
asked Jan 2 at 11:11
Michele Da RosMichele Da Ros
418215
418215
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Found it.
The trick was to use MultipartFormDataOutput and to set the filename when adding a part
MultipartFormDataOutput multipartOutput = new MultipartFormDataOutput();
multipartOutput.addFormData("uploaded file", file, MediaType.APPLICATION_OCTET_STREAM_TYPE, "file.txt");
answered Jan 3 at 9:53
Michele Da RosMichele Da Ros
418215
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Found it.
The trick was to use MultipartFormDataOutput and to set the filename when adding a part
MultipartFormDataOutput multipartOutput = new MultipartFormDataOutput();
multipartOutput.addFormData("uploaded file", file, MediaType.APPLICATION_OCTET_STREAM_TYPE, "file.txt");
answered Jan 3 at 9:53
Michele Da RosMichele Da Ros
418215
add a comment |
Found it.
The trick was to use MultipartFormDataOutput and to set the filename when adding a part
MultipartFormDataOutput multipartOutput = new MultipartFormDataOutput();
multipartOutput.addFormData("uploaded file", file, MediaType.APPLICATION_OCTET_STREAM_TYPE, "file.txt");
answered Jan 3 at 9:53
Michele Da RosMichele Da Ros
418215
add a comment |
Found it.
The trick was to use MultipartFormDataOutput and to set the filename when adding a part
MultipartFormDataOutput multipartOutput = new MultipartFormDataOutput();
multipartOutput.addFormData("uploaded file", file, MediaType.APPLICATION_OCTET_STREAM_TYPE, "file.txt");
answered Jan 3 at 9:53
Michele Da RosMichele Da Ros
418215
Found it.
The trick was to use MultipartFormDataOutput and to set the filename when adding a part
MultipartFormDataOutput multipartOutput = new MultipartFormDataOutput();
multipartOutput.addFormData("uploaded file", file, MediaType.APPLICATION_OCTET_STREAM_TYPE, "file.txt");
answered Jan 3 at 9:53
Michele Da RosMichele Da Ros
418215
answered Jan 3 at 9:53
Michele Da RosMichele Da Ros
418215
answered Jan 3 at 9:53
Michele Da RosMichele Da Ros
418215
answered Jan 3 at 9:53
Michele Da RosMichele Da Ros
418215
418215
add a comment |
add a comment |
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