Question

以下是我的REST API代码：

 @RequestMapping(value = "/get", method = RequestMethod.POST, produces = { "application/json" })
      @ApiOperation(value = "get data by key.", notes = "return json string value.")
      public JsonObjectResponse<String> get(
          @ApiParam(required = true, name = "regionName", value = "region name") @RequestParam("regionName") String regionName,
          @ApiParam(required = true, name = "key", value = "region key,Default is uuid") @RequestParam("key") String key) throws UnsupportedEncodingException
      {
        JsonObjectResponse<String> jr = new JsonObjectResponse<String>();
        //key = decodeJsonString(key);  // added for junit
        String val = adfService.onPath(regionName).get(key);
        jr.setState(StateCode.SUCCESS);
        jr.setData(JsonObject.create().append(key,val).toJson());

        return jr;
      }

我正在尝试传递参数：

regionName=/fusion/table1&key={"fusionTbl1DetailNo":"fusionNo001","pk":"PK0001"}

如果我通过swagger-ui调用它，它会这样调用：

http://localhost:8080/service/basic/get?regionName=%2Ffusion%2Ftable1&key=%7B%22fusionTbl1DetailNo%22%3A%22fusionNo001%22%2C%22pk%22%3A%22PK0001%22%7D&token=8652493a-4147-43f4-af3a-bcb117fb7d42`

它对参数进行了编码，这些参数也可以在服务器端自动解码。

当我想为此API添加测试用例时，我使用restTemplate.exchange方法，代码如下：

UriComponentsBuilder builder = UriComponentsBuilder.fromHttpUrl(url);
for (Entry<String, String> entry : queryParamMap.entrySet()) {
    builder.queryParam(entry.getKey(), entry.getValue());
}
if (uriParamMap != null) {

    url = builder.buildAndExpand(uriParamMap).toUriString();
} else {
    url = builder.toUriString();
}
if (StringUtils.isEmpty(requestBody)) {
    if (bodyParamMap != null) {
        requestBody = parseMapToParams(bodyParamMap);
    } else {
        requestBody = "";
    }
}
HttpHeaders headers = new HttpHeaders();
MediaType mediaType = new MediaType("application", "json", Charset.forName("UTF-8"));
headers.setContentType(MediaType.APPLICATION_JSON_UTF8);
// headers.add(HttpHeaders.CONTENT_TYPE, "application/json");
// headers.add("Accept", "application/json");
// headers.set(HttpHeaders.ACCEPT, "application/json");
headers.setAccept(Arrays.asList(MediaType.APPLICATION_JSON));
// headers.add("Accept-Encoding", "gzip, deflate, br");
headers.set("Accept-Charset", "utf-8");
headers.set("Accept-Encoding", "gzip");
headers.add("Accept-Language", "zh-CN,zh;q=0.8,en;q=0.6");
headers.add("User-Agent",
        "Mozilla/5.0 (Windows NT 6.1; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/58.0.3029.110 Safari/537.36");

headers.add(TestBase.TOKEN_HEADER, TestBase.getTokenId());
HttpEntity<String> request = new HttpEntity<>(body, headers);
restTemplate.getMessageConverters().add(0, new StringHttpMessageConverter(Charset.forName("UTF-8")));
ResponseEntity<String> response = restTemplate.exchange(url, httpMethod, request, String.class);
localresponse.set(response);
System.out.println("response:" + response);
return response;

我使用UriComponentsBuilder附加参数，它会将网址格式化为

http://localhost:8080/service/basic/get?regionName=/fusion/table1&key=%7B%22fusionTbl1DetailNo%22:%22fusionNo001%22,%22pk%22:%22PK0001%22%7D

exchange

。但是，当服务器端收到呼叫时，它没有解码参数key，它的值仍然是：

%7B%22fusionTbl1DetailNo%22:%22fusionNo001%22,%22pk%22:%22PK0001%22%7D

为什么？我比较了swagger调用的标题设置，添加了额外的设置，没有效果:(。

Answer 1

尝试以下内容：

ResponseEntity<String> res = restTemplate.exchange("http://localhost:8080/service/basic/get?regionName={arg1}&key={arg2}", HttpMethod.POST, null, String.class,"/fusion/table1", "{\"fusionTbl1DetailNo\":\"fusionNo001\",\"pk\":\"PK0001\"}");

arg1和arg2将替换为

"/fusion/table1"和"{\"fusionTbl1DetailNo\":\"fusionNo001\",\"pk\":\"PK0001\"}"

我在null中发送requestEntity，因为没有请求正文和请求参数在uriVariables中。

Answer 2

RestTemplate上的Spring文档读取：

对于每个HTTP方法，有三种变体：两种接受URI 模板字符串和URI变量（数组或映射），而第三个接受一个URI。请注意，对于URI模板，假设编码为必要的，例如restTemplate.getForObject（＆＃34; http://example.com/hotel list＆＃34;）成为＆＃34; http://example.com/hotel%20list＆＃34;。这也意味着 URI模板或URI变量已经编码，双重编码会发生，

看起来你正在使用一个带有URI模板字符串的RestTemplate交换方法，因此你不应该对url-string进行编码。

首先在

上编码url-string

builder.toUriString()

然后又在交换电话上。所以问题似乎是客户端的双重编码，而不是服务器端的解码不足

Answer 3

而不是toUriString()使用，UriComponentsBuilder.fromUriString(url).queryParam("name","John Doe").build().toString();

resttemplate.exchange传递的param未在服务端自动解码

3 个答案: