Java – how do I receive multiple files in InputStream and process them accordingly?
•
Java
I want to receive multiple files uploaded from the client I uploaded multiple files and requested my server (Java) using Jax - RS (Jersey)
I have the following code,
@POST
@Consumes(MediaType.MULTIPART_FORM_DATA)
public void upload(@Context UriInfo uriInfo,@FormDataParam("file") final InputStream is,@FormDataParam("file") final FormDataContentDisposition detail) {
FileOutputStream os = new FileOutputStream("Path/to/save/" + appropriatefileName);
byte[] buffer = new byte[1024];
int length;
while ((length = is.read(buffer)) > 0) {
os.write(buffer,length);
}
}
How to write files separately on the server side uploaded by the client
For example I uploaded my_ File. txt,My_ File. PNG,My_ File. Doc and other documents I need to write my on the server side_ File. txt,My_ File. Doc
How can I do this?
Solution
You can try something like this:
@POST
@Consumes(MediaType.MULTIPART_FORM_DATA)
public void upload(FormDataMultiPart formParams)
{
Map<String,List<FormDataBodyPart>> fieldsByName = formParams.getFields();
// Usually each value in fieldsByName will be a list of length 1.
// Assuming each field in the form is a file,just loop through them.
for (List<FormDataBodyPart> fields : fieldsByName.values())
{
for (FormDataBodyPart field : fields)
{
InputStream is = field.getEntityAs(InputStream.class);
String fileName = field.getName();
// TODO: SAVE FILE HERE
// if you want media type for validation,it's field.getMediaType()
}
}
}
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