small correction on @PetrosTsialiamanis post , new File( multipart.getOriginalFilename()) this will create file in server location where sometime you will face write permission issues for the user, its not always possible to give write permission to every user who perform action.System.getProperty("java.io.tmpdir") will create temp directory where Trending. This guide explains how to use the RESTEasy REST Client to send multipart REST requests, typically to upload documents. 2. It sends the form data to server in multiple parts because of large size of file. Introduction. DataWeave can read input data as a whole in-memory, in indexed fashion, and for some data formats, part-by-part by streaming the input. In the details panel, click Create table add_box.. On the Create table page, in the Source section:. For example: After configuring all the things click on send and see out put like this -- see image. FileUploadExceptionAdvice handles exception when Step 7- Run the application and use Postman to test Web API.If you are not aware about Postman, click here, otherwise see in the image how to configure Postman to test Web API. I want to upload a file from input to a server in multipart/form-data. I want to upload a file from input to a server in multipart/form-data. I googled and implemented all suggested solutions but none works, like: solutions suggested here, and so on.. First: headers: { 'Content-Type': undefined }, Which results in e.g. DataWeave can read input data as a whole in-memory, in indexed fashion, and for some data formats, part-by-part by streaming the input. Multipart form data: The ENCTYPE attribute of