Question

我有以下curl表达式：

curl --data 'api_key=API_Key' --data-urlencode 'event=[{"user_id":"12345", "event_type":"buy_song"}]' https://someapi

应转换为RestTemplate.postForEntity调用。我是这样转换的：

HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_FORM_URLENCODED);

MultiValueMap<String, String> params = new LinkedMultiValueMap<>();
params.add("api_key", "API_Key");
params.add("event", URLEncoder.encode(objectMapper.writeValueAsString(Collections.singletonList(e)), "UTF-8"));

// send
HttpEntity<MultiValueMap<String, String>> request = new HttpEntity<>(params, headers);
ResponseEntity<String> response = restTemplate.postForEntity("https://someapi", request, String.class);

服务器返回 400错误的请求

我确认杰克逊的对象映射器正确地序列化了对象objectMapper.writeValueAsString(Collections.singletonList(e))

我怀疑我无法从RestTemplate中的示例curl中正确处理--data和--data-urlencode的混合。

请问我在做什么错？

Answer 1

  // org.apache.commons.collections.map.HashedMap     
  HashedMap requestBody = new HashedMap();
      requestBody.put("api_key", "API_Key");
      requestBody.put("event", URLEncoder.encode(objectMapper.writeValueAsString(Collections.singletonList(e)), "UTF-8"));   

        HttpHeaders headers = new HttpHeaders();
        headers.setContentType(MediaType.APPLICATION_FORM_URLENCODED);
       String jsonBody = new ObjectMapper().writeValueAsString(requestBody);
       HttpEntity<String> entity = new HttpEntity<>(jsonBody, headers);

        ResponseEntity<String> response = restTemplate.postForEntity("https://someapi", entity, String.class);

Answer 2

对我来说很好：

这是客户端和请求的代码，请忽略请求签名

curl -X GET \
  http://localhost:8080/article/so \
  -H 'cache-control: no-cache' \
  -d 'api_key=API_Key&event=%5B%7B%22user_id%22%3A%2212345%22%2C%20%22event_type%22%3A%22buy_song%22%7D%5D'

控制器：首次请求将触发您的代码：

// This is just to trigger, you can ignore it.

    @RequestMapping(value = "/article/so", method = RequestMethod.GET)
    public void addArticle1() throws UnsupportedEncodingException, JsonProcessingException {
        articleServiceImpl.test();
    }

以上请求将转到与您的代码相同的服务层。


    public void test() throws JsonProcessingException, UnsupportedEncodingException {
        HttpHeaders headers = new HttpHeaders();
        headers.setContentType(MediaType.APPLICATION_FORM_URLENCODED);

        MultiValueMap<String, String> params = new LinkedMultiValueMap<>();
        Event e = new Event();
        e.setEvent_type("buy_song");
        e.setUser_id("12345");

        params.add("api_key", "API_Key");
        String encode = URLEncoder.encode(objectMapper.writeValueAsString(Collections.singletonList(e)), "UTF-8");
        params.add("event", encode);

        HttpEntity<MultiValueMap<String, String>> request = new HttpEntity<>(params, headers);
        ResponseEntity<String> response = restTemplate.postForEntity("http://localhost:8080/article/so", request, String.class);
        System.out.println(response);//<200,[Content-Length:"0", Date:"Fri, 31 May 2019 09:26:24 GMT"]>
    }

然后再次使用控制器，只是检查控制器是否从调用方接受String对象或Event对象。在这里，事件作为URLEncodedString传递，我在这里将其作为输出。

    @RequestMapping(value = "/article/so", method = RequestMethod.POST, consumes = MediaType.APPLICATION_FORM_URLENCODED_VALUE)
    public void addArticle2(@RequestParam String api_key,@RequestParam String event) {
        System.out.println(api_key); // API_Key
        System.out.println(event); // %5B%7B%22user_id%22%3A%2212345%22%2C%22event_type%22%3A%22buy_song%22%7D%5D
    }

Answer 3

我认为唯一的问题是MediaType，您发送的数据不是数据（APPLICATION_FORM_URLENCODED）

它是json数据，因此您需要使用MediaType.APPLICATION_JSON这样的东西

HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_JSON);

// converting form variable to Map
MultiValueMap<String, String> map= new LinkedMultiValueMap<>();
map.add("api_key", "API_Key");
map.add("event", URLEncoder.encode(objectMapper.writeValueAsString(Collections.singletonList(e)), "UTF-8"));

// finally build Request
HttpEntity<MultiValueMap<String, String>> request = new HttpEntity<>(map, headers);

ResponseEntity<String> response = restTemplate.postForEntity(
  apiUrl, request , String.class);

有关{RestTemplate的更多信息，请参见this

将curl命令转换成RestTemplate

3 个答案: