在JSON中的RestTemplate中获取GET请求

时间:2017-02-09 17:10:11

标签: java spring http get resttemplate

对于一个看似超级直截了当的问题,已经困扰了几天:

我在application/json中使用RestTemplate进行简单的GET请求,但我一直在

org.springframework.web.client.HttpClientErrorException: 400 Bad Request
    at org.springframework.web.client.DefaultResponseErrorHandler.handleError(DefaultResponseErrorHandler.java:91)
    at org.springframework.web.client.RestTemplate.handleResponse(RestTemplate.java:636)
    at org.springframework.web.client.RestTemplate.doExecute(RestTemplate.java:592)
    at org.springframework.web.client.RestTemplate.execute(RestTemplate.java:552)
    at org.springframework.web.client.RestTemplate.exchange(RestTemplate.java:470)

我做了研究并遵循了这个tutorial,我也查看了这个POST request via RestTemplate in JSON的解决方案。但是他们都没有帮助,这是我的代码:

RestTemplate restTemplate = new RestTemplate();    
HttpHeaders requestHeaders = new HttpHeaders();    
requestHeaders.setContentType(MediaType.APPLICATION_JSON);
HttpEntity<?> requestEntity = new HttpEntity<Object>(requestHeaders);
restTemplate.exchange(endpoint, HttpMethod.GET, requestEntity, String.class);

endpointhttp://localhost:8080/api/v1/items?itemIds=" + URLEncoder.encode(itemIds, "UTF-8"),在邮差中效果很好。 itemIds是逗号分隔的列表,如下所示:

5400028914,5400029138,5400029138,5400029138,5400029138,5400028401,5400028918,5400028076,5400028726

我也尝试使用getForObject,如下所示:

String result = restTemplate.getForObject(endpoint, String.class);

这给了我这个错误:

org.springframework.web.client.HttpClientErrorException: 415 Unsupported Media Type

我不确定我错过了什么或做错了什么,但是相同的端点在Postman上完美运行,但唯一的区别是我在Postman应用程序中添加了Content-Type标头。

这是我的邮递员要求:

GET /api/v1/items?itemIds=abc%2cdef%2cghi HTTP/1.1 Host: localhost:8080 Connection: keep-alive Postman-Token: 84790e06-86aa-fa8a-1047-238d6c931a68 Cache-Control: no-cache User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_11_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/56.0.2924.87 Safari/537.36 Content-Type: application/json Accept: */* Accept-Encoding: gzip, deflate, sdch Accept-Language: en-US,en;q=0.8,zh-CN;q=0.6,zh;q=0.4

那么如果上面的代码错误,如何使用RestTemplate正确设置内容类型?

另一次深度探索,我已经启动Wireshark来捕获两个HTTP请求,以下是截图:

邮递员的要求:request from Postman

来自我的Java代码的请求:request from Java program

我仍然不明白为什么我的Java程序中的那个抛出400而Postman的那个工作正常。

非常感谢。

2 个答案:

答案 0 :(得分:1)

好吧,最终,我的一位同事帮我弄明白了为什么,不管你信不信,这很简单:

endpoint是这样的: "http:localhost:8080/api/v1/items?itemIds=" + URLEncoder.encode(itemIds, "UTF-8");

但是,它应该是"http:localhost:8080/api/v1/items?itemIds=" + itemIds;

itemIds只是一个以逗号分隔的列表。

URLEncoder编码经过&#34; UTF-8&#34;架构,这个以逗号分隔的列表变为itemIds=5400028914%2C5400029138%2C5400029138%2C5400029138%2C5400029138%2C5400028401%2C5400028918%2C5400028076

来自

itemIds=5400028914,5400029138,5400029138,5400029138,5400029138,5400028401,5400028918,5400028076,5400028726

使用RestTemplate时,我们不需要使用URLEncoder来编码URL,有人可以帮我加深理解吗?

谢谢!

答案 1 :(得分:0)

您不需要为GET方法设置“requestEntity”,请尝试使用以下代码:

public ItemInfo getItemInfo() throws Exception {
    String url =
            "http://localhost:8080/api/v1/items?itemIds=abc";
    ObjectMapper objectMapper = new ObjectMapper();     
    ResponseEntity<String> response = restTemplate.exchange(url, HttpMethod.GET, null, String.class);
    String responseBody = response.getBody();
    handlerError(response, url);
    try {
        return objectMapper.readValue(responseBody, ItemInfo.class);
    } catch (IOException exception) {
        LOGGER.error("failed to send REST request", exception);
        throw new Exception(ErrorCode.NOT_AVAILABLE, url);
    }
}

private void handlerError(final ResponseEntity<String> response, final String url) throws Exception {
    String responseBody = response.getBody();
    try {
        if (RestUtil.isError(response.getStatusCode())) {
            ObjectMapper objectMapper = new ObjectMapper(); 
            MyInfoError myInfoError = objectMapper.readValue(responseBody, MyInfoError.class);
            throw new Exception(infoErreur, ErrorCode.UNKNOWN_CODE_ERROR);
        } else if (RestUtil.isNotFound(response.getStatusCode())) {
            throw new Exception(ErrorCode.NOT_AVAILABLE, "MyService");
        }
    } catch (IOException exception) {
        LOGGER.error("failed to send REST request", exception);
        throw new Exception(ErrorCode.NOT_AVAILABLE, url);
    }
}

我把它设为NULL因为GET方法没有发送任何JSON请求正文/标题:

restTemplate.exchange(url, HttpMethod.GET, null, String.class);

或者:将您的标题放在GET方法中,如下所示:

RestTemplate restTemplate = new RestTemplate();
HttpHeaders headers = new HttpHeaders();
headers.setAccept(Arrays.asList(MediaType.APPLICATION_JSON));
HttpEntity<?> requestEntity = new HttpEntity<>(headers);
ResponseEntity<String> response = restTemplate.exchange(url, HttpMethod.GET, requestEntity, String.class);

对于POST的示例,您需要设置requestEntity,如下所示:

ItemFormRequest request =
            new ItemFormRequest(1,"item no", "item name");
HttpEntity<ItemFormRequest> requestEntity = new HttpEntity<>(request);
        ResponseEntity<String> response = restTemplate.exchange(url, HttpMethod.POST, requestEntity, String.class);
        String responseBody = response.getBody();

希望这有帮助:)