如何将 POST 请求中包含文件的 MultiValueMap 发送到 Spring Rest Web 服务？

Question

我想使用restTemplate 发送一个POST 请求，其中包括一个文件和另一个自定义对象。我已经尝试过下面的代码，但没有让它工作，因为它缺少HtttpMessageConverterforjava.io.File

MultiValueMap 是否是在一个请求中发送两个对象的正确选择？如果没有，有什么建议吗？如果是，那么如何解决消息转换器问题？

客户端（从这里和这里采用）：

    RestTemplate rest = new RestTemplate();
    HttpMessageConverter formHttpMessageConverter = new FormHttpMessageConverter();
    HttpMessageConverter stringHttpMessageConverternew = new StringHttpMessageConverter();
    List<HttpMessageConverter<?>> conv = new ArrayList<HttpMessageConverter<?>> ();
    conv.add(formHttpMessageConverter);
    conv.add(stringHttpMessageConverternew);

    rest.setMessageConverters(conv);
    MultiValueMap<String, Object> map = new LinkedMultiValueMap<String, Object>();
    File f = new File("testFile.txt");
    map.add("file", f);
    NewObject a = new NewObject("123", "xxx", "xxx",
            "blahblah", "5432");
    map.add("newObject", a);
    String result = rest.postForObject("http://localhost:8080/test/upload", map, String.class);
    System.out.println(result);

控制器：

@RequestMapping(value="/upload", method=RequestMethod.POST)
public @ResponseBody String uploadArtefact(@RequestBody MultiValueMap<String, Object> o)
{
    logger.info("tesssting: " + o);
    File x = (File) o.get("file").get(0);
    logger.info("File name " + x.getName());
    NewObject a = (NewObject) o.get("newObject").get(0);
    logger.info("New Object: " + a);
    return "finished";


}

这是我在客户端遇到的错误：

Exception in thread "main" org.springframework.http.converter.HttpMessageNotWritableException: Could not write request: no suitable HttpMessageConverter found for request type [java.io.File]
at org.springframework.http.converter.FormHttpMessageConverter.writePart(FormHttpMessageConverter.java:317)
at org.springframework.http.converter.FormHttpMessageConverter.writeParts(FormHttpMessageConverter.java:275)
at org.springframework.http.converter.FormHttpMessageConverter.writeMultipart(FormHttpMessageConverter.java:264)
at org.springframework.http.converter.FormHttpMessageConverter.write(FormHttpMessageConverter.java:204)
at org.springframework.http.converter.FormHttpMessageConverter.write(FormHttpMessageConverter.java:72)
at org.springframework.web.client.RestTemplate$HttpEntityRequestCallback.doWithRequest(RestTemplate.java:751)
at org.springframework.web.client.RestTemplate.doExecute(RestTemplate.java:540)
at org.springframework.web.client.RestTemplate.execute(RestTemplate.java:503)
at org.springframework.web.client.RestTemplate.postForObject(RestTemplate.java:331)
at uk.ac.ncl.rest.Test.main(Test.java:47)

这两行记录在服务器端：

2015-02-24 16:44:14.772  WARN 38776 --- [nio-8080-exec-1] .c.j.MappingJackson2HttpMessageConverter : Failed to evaluate deserialization for type [map type; class org.springframework.util.MultiValueMap, [simple type, class java.lang.String] -> [collection type; class java.util.List, contains [simple type, class java.lang.Object]]]: com.fasterxml.jackson.databind.JsonMappingException: Can not find a deserializer for non-concrete Map type [map type; class org.springframework.util.MultiValueMap, [simple type, class java.lang.String] -> [collection type; class java.util.List, contains [simple type, class java.lang.Object]]]
2015-02-24 16:44:14.772  WARN 38776 --- [nio-8080-exec-1] .c.j.MappingJackson2HttpMessageConverter : Failed to evaluate deserialization for type [map type; class org.springframework.util.MultiValueMap, [simple type, class java.lang.Object] -> [collection type; class java.util.List, contains [simple type, class java.lang.Object]]]: com.fasterxml.jackson.databind.JsonMappingException: Can not find a deserializer for non-concrete Map type [map type; class org.springframework.util.MultiValueMap, [simple type, class java.lang.Object] -> [collection type; class java.util.List, contains [simple type, class java.lang.Object]]]

Answer 1

我在这个错误上花了很多时间！

FormHttpMessageConverter 有 3 个部分转换器：

1. ByteArrayHttpMessageConverter - 写入 byte[].class
2. StringHttpMessageConverter - String.class
3. ResourceHttpMessageConverter - org.springframework.core.io.Resource.class - 这是技巧

您应该将其作为资源，而不是放置文件本身：

File file = new File("some_path");
Resource resource = new FileSystemResource(file);
map.add("file", resource);

FileSystemResource 来自包 org.springframework.core.io