Spring RestTemplate GET with parameters

Question

I have to make a REST call that includes custom headers and query parameters. I set my HttpEntity with just the headers (no body), and I use the RestTemplate.exchange() method as follows:

HttpHeaders headers = new HttpHeaders();
headers.set("Accept", "application/json");

Map<String, String> params = new HashMap<String, String>();
params.put("msisdn", msisdn);
params.put("email", email);
params.put("clientVersion", clientVersion);
params.put("clientType", clientType);
params.put("issuerName", issuerName);
params.put("applicationName", applicationName);

HttpEntity entity = new HttpEntity(headers);

HttpEntity<String> response = restTemplate.exchange(url, HttpMethod.GET, entity, String.class, params);

This fails at the client end with the dispatcher servlet being unable to resolve the request to a handler. Having debugged it, it looks like the request parameters are not being sent.

When I do a an exchange with a POST using a request body and no query parameters it works just fine.

Does anyone have any ideas?

Answer 1

If you pass non-parametrized params for RestTemplate, you'll have one Metrics for everyone single different URL that you pass, considering the parameters. You would like to use parametrized urls:

http://my-url/action?param1={param1}&param2={param2}

instead of

http://my-url/action?param1=XXXX&param2=YYYY

The second case is what you get by using UriComponentsBuilder class.

One way to implement the first behavior is the following:

Map<String, Object> params = new HashMap<>();
params.put("param1", "XXXX");
params.put("param2", "YYYY");

String url = "http://my-url/action?%s";

String parametrizedArgs = params.keySet().stream().map(k ->
    String.format("%s={%s}", k, k)
).collect(Collectors.joining("&"));

HttpHeaders headers = new HttpHeaders();
headers.set("Accept", MediaType.APPLICATION_JSON_VALUE);
HttpEntity<String> entity = new HttpEntity<>(headers);

restTemplate.exchange(String.format(url, parametrizedArgs), HttpMethod.GET, entity, String.class, params);

Answer 2

I take different approach, you may agree or not but I want to control from .properties file instead of compiled Java code

Inside application.properties file

endpoint.url = https://yourHost/resource?requestParam1={0}&requestParam2={1}

Java code goes here, you can write if or switch condition to find out if endpoint URL in .properties file has @PathVariable (contains {}) or @RequestParam (yourURL?key=value) etc... then invoke method accordingly... that way its dynamic and not need to code change in future one stop shop...

I'm trying to give more of idea than actual code here ...try to write generic method each for @RequestParam, and @PathVariable etc... then call accordingly when needed

  @Value("${endpoint.url}")
  private String endpointURL;
  // you can use variable args feature in Java
  public String requestParamMethodNameHere(String value1, String value2) {
    RestTemplate restTemplate = new RestTemplate();
    restTemplate
           .getMessageConverters()
           .add(new MappingJackson2HttpMessageConverter());

    HttpHeaders headers = new HttpHeaders();
    headers.set("Accept", MediaType.APPLICATION_JSON_VALUE);
    HttpEntity<String> entity = new HttpEntity<>(headers);

    try {
      String formatted_URL = MessageFormat.format(endpointURL, value1, value2);
      ResponseEntity<String> response = restTemplate.exchange(
                    formatted_URL ,
                    HttpMethod.GET,
                    entity,
                    String.class);
     return response.getBody();
    } catch (Exception e) { e.printStackTrace(); }

Answer 3

To easily manipulate URLs / path / params / etc., you can use Spring's UriComponentsBuilder class. It's cleaner that manually concatenating strings and it takes care of the URL encoding for you:

HttpHeaders headers = new HttpHeaders();
headers.set("Accept", MediaType.APPLICATION_JSON_VALUE);

UriComponentsBuilder builder = UriComponentsBuilder.fromHttpUrl(url)
        .queryParam("msisdn", msisdn)
        .queryParam("email", email)
        .queryParam("clientVersion", clientVersion)
        .queryParam("clientType", clientType)
        .queryParam("issuerName", issuerName)
        .queryParam("applicationName", applicationName);

HttpEntity<?> entity = new HttpEntity<>(headers);

HttpEntity<String> response = restTemplate.exchange(
        builder.toUriString(), 
        HttpMethod.GET, 
        entity, 
        String.class);

Answer 4

OK, so I'm being an idiot and I'm confusing query parameters with url parameters. I was kinda hoping there would be a nicer way to populate my query parameters rather than an ugly concatenated String but there we are. It's simply a case of build the URL with the correct parameters. If you pass it as a String Spring will also take care of the encoding for you.

Answer 5

I was attempting something similar, and the RoboSpice example helped me work it out:

HttpHeaders headers = new HttpHeaders();
headers.set("Accept", "application/json");

HttpEntity<String> request = new HttpEntity<>(input, createHeader());

String url = "http://awesomesite.org";
Uri.Builder uriBuilder = Uri.parse(url).buildUpon();
uriBuilder.appendQueryParameter(key, value);
uriBuilder.appendQueryParameter(key, value);
...

String url = uriBuilder.build().toString();

HttpEntity<String> response = restTemplate.exchange(url, HttpMethod.GET, request , String.class);

Answer 6

The uriVariables are also expanded in the query string. For example, the following call will expand values for both, account and name:

restTemplate.exchange("http://my-rest-url.org/rest/account/{account}?name={name}",
    HttpMethod.GET,
    httpEntity,
    clazz,
    "my-account",
    "my-name"
);

so the actual request url will be

http://my-rest-url.org/rest/account/my-account?name=my-name

Look at HierarchicalUriComponents.expandInternal(UriTemplateVariables) for more details. Version of Spring is 3.1.3.