JsonMappingException：无法构造CommonsMultipartFile的实例 [英] JsonMappingException: Can not construct instance of CommonsMultipartFile

问题描述

我使用Spring-MVC 3，在我的应用程序中，我发送一些信息
有多个附件，每个文件都有一个标题，Id等等。所以，我做了一个DTO如下

I using Spring-MVC 3, and in my application, I am sending some information with multiple attachments and each one this files have one title, Id etc. So, I made one DTO as follows

public class MyDTO {

Long id;

Integer age;

MultipartFile infoFile;

// getter setter

我只是创建一个 JSON 根据我的 JS 文件中的上述DTO类对象。

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

这里是我的控制器映射：

@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();
        }

但我收到以下错误：

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"])

推荐答案

其实我找到了自己的答案。我们无法直接在JSON对象中发送文件。 文件对象不保存文件，它保存文件的路径，即。 C：/hi.txt。如果这就是我们在JSON中放入的内容，它将产生

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

{文件：C：/hi.txt}

它不包含文件内容。所以我们不妨直接放置文件路径

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");

如果您尝试使用JSON上传文件，一种方法是从中读取字节数Java 7的NIO文件

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对这些字节进行编码，并将它们作为字符串写入JSONObject中。使用Apache Commons Codec

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));

有 94个Unicode 字符可以是根据JSON规范表示为一个字节（如果您的JSON以UTF-8格式传输）。

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).

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