如何通过RestTemplate客户端和服务器将映像上传到Server()

时间:2016-02-15 12:32:23

标签: java spring-mvc resttemplate

我正在使用RestTemplate客户端代码并希望在服务器端访问Image。我想在自定义Tomcat目录上传图像,我收到错误:

Caused by: org.springframework.web.client.RestClientException: Could not write request: no suitable HttpMessageConverter found for request type [org.apache.http.entity.mime.MultipartEntity] and content type [multipart/form-data]

RestTemplate客户端代码:

public String saveCompanylogo(File file){
        String url = COMPANY_URL+ "/saveCompanyLogo";
        MultipartEntity multiPartEntity = new MultipartEntity ();
        FileBody fileBody = new FileBody(file) ;
        //Prepare payload
        multiPartEntity.addPart("file", fileBody) ;
        HttpHeaders headers = new HttpHeaders();
        headers.setContentType(MediaType.MULTIPART_FORM_DATA);
        HttpEntity<MultipartEntity> entity = new HttpEntity<MultipartEntity>    (multiPartEntity, headers);
        ResponseEntity<String> exchange = restTemplate.exchange(url,     HttpMethod.POST, entity,  new ParameterizedTypeReference<String>() {
        });
        return exchange.getBody();
    }

我的服务器端(Controller)代码是:

@RequestMapping(method = POST, value = "/saveCompanyLogo")
    @Consumes("multipart/form-data")
    public String saveCompanylogo(@RequestParam("file") MultipartFile file)         {
      System.out.println(""+file);   
     //Todo coding
     return "stringData";
    }

1 个答案:

答案 0 :(得分:1)

我使用FileSystemResource代替FileBody。这是我的代码:

restTemplate.getMessageConverters().add(new ByteArrayHttpMessageConverter());

Map params = new LinkedMultiValueMap();
params.add("file", new FileSystemResource(file));
HttpHeaders httpHeaders = new HttpHeaders();
httpHeaders.setContentType(MediaType.MULTIPART_FORM_DATA);
HttpEntity requestEntity = new HttpEntity<>(params, httpHeaders);
restTemplate.exchange(url, HttpMethod.POST, requestEntity, String.class);

在服务器端(我使用Jersey而不是MVC,但我认为无关紧要):

@RequestMapping(method = POST, value = "/saveCompanyLogo")
@Consumes("multipart/form-data")
public String saveCompanylogo(@RequestParam("file") InputStream file) {
    //DO SOMETHING
}

希望有所帮助