I am having a Rest URL with form data which i succesfully execute from my REST Client POST MAN.How to do this from a java program ? How to pass the attached files programtically. The snapshot of the rest call is as follows.
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Here's a [Jersey option](http://stackoverflow.com/a/27614403/2587435) – Paul Samsotha Sep 11 '15 at 10:52
1 Answers
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check the below code example.
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.FormParam;
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,
@FormDataParam("path") String path) {
// Path format //10.217.14.97/Installables/uploaded/
System.out.println("path::"+path);
String uploadedFileLocation = path
+ 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();
}
}

Ram
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