JsonMappingException: Can not construct instance of CommonsMultipartFile

Question

I am using Spring-MVC 3.0, and in my application, I am sending some information with multiple attachments, and each one of these files have title, Id etc. So, I made one DTO as follows

public class MyDTO {

Long id;

Integer age;

MultipartFile infoFile;

// getter setter

I am just creating one JSON object according to the above DTO class in my JS file.

Here is my Controller mapping:

@RequestMapping(value = "/saveInfo", method = RequestMethod.POST)
public @ResponseBody String saveInfo(
       @RequestParam(value = "data", required = true) String stdData,
       @RequestParam(value = "fileData", required = false) MultipartFile[] files,
       HttpSession session,HttpServletRequest request) {
    
       MyDTO dto;
       try {
                dto = mapper.readValue(stdData, new TypeReference<MyDTO>() {});  
        } catch (JsonParseException e) {
                e.printStackTrace();
        } catch (JsonMappingException e) {
                e.printStackTrace();
        } catch (IOException e) {
                e.printStackTrace();
        }

But I am getting following errors:

org.codehaus.jackson.map.JsonMappingException: Can not construct instance of    org.springframework.web.multipart.commons.CommonsMultipartFile, 
problem: no suitable creator method found to deserialize from JSON String
at [Source: java.io.StringReader@19747c9; line: 1, column: 336] (through reference chain: com.avi.dto.MyDTO["hbvFile"])

It probably wants a no-arg constructor, and `CommonsMultipartFile` does not have one. — Oleg Estekhin, Jul 08 '14 at 07:31

score 4 · Accepted Answer · answered Jul 14 '14 at 10:15

Actually I find the answer for myself. We can't send file directly in JSON object. A File object doesn't hold a file, it holds the path to the file, ie. C:/hi.txt. If that's what we put in our JSON, it'll produce

{"File" : "C:/hi.txt"}

It won't contain the file content. So we might as well just put the file path directly

JSONObject my_data = new JSONObject();
my_data.put("User", "Avi");
my_data.put("Date", "22-07-2013");
my_data.put("File", "C:/hi.txt");

If you're trying to do a file upload with JSON, one way is to read the bytes from the file with Java 7's NIO

byte[] bytes = Files.readAllBytes(file_upload .toPath());

Base64 encode those bytes and write them as a String in the JSONObject. Using Apache Commons Codec

Base64.encodeBase64(bytes);
my_data.put("File", new String(bytes));

There are 94 Unicode characters which can be represented as one byte according to the JSON spec (if your JSON is transmitted as UTF-8).

JsonMappingException: Can not construct instance of CommonsMultipartFile

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