来自 RestTemplate java 的 azure 逻辑应用程序的调用请求触发器 url

Question

我有带有请求触发器的 Azure 逻辑应用程序。 我想从我的 java 应用程序中触发这个逻辑应用程序。 所以，我试图从我的 java API 调用请求触发器 url。

如果我使用 DefaultHttpClient 但在 java 中使用 RestTemplate 调用它时得到 401，它工作正常。

DefaultHttpClient 代码：

try {
    DefaultHttpClient httpClient = new DefaultHttpClient();

    HttpGet getRequest = new HttpGet(
        "{url of azure logic app trigger}");

    //StringEntity input = new StringEntity("{\"qty\":100,\"name\":\"iPad 4\"}");
    //input.setContentType("application/json");
    //postRequest.setEntity(input);

    HttpResponse response = httpClient.execute(getRequest);

    BufferedReader br = new BufferedReader(
                    new InputStreamReader((response.getEntity().getContent())));

    String output;
    System.out.println("Output from Server .... \n");
    while ((output = br.readLine()) != null) {
        System.out.println(output);
    }


    httpClient.getConnectionManager().shutdown();
    return("Success");
} catch (ClientProtocolException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
    return("Error");
} catch (UnsupportedOperationException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
    return("Error");
} catch (IOException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
    return("Error");
}

休息模板代码

@Service
public class SampleService {
@Autowired HttpClientService<String, String> httpClientService;
public String callURL() {
ResponseErrorHandler responseErrorHandler = new ResponseErrorHandler() {

    @Override
    public boolean hasError(ClientHttpResponse response) throws IOException {
        System.out.print(response.toString());
        return false;
    }

    @Override
    public void handleError(ClientHttpResponse response) throws IOException {
        // TODO Auto-generated method stub
    }
};
UriComponentsBuilder builder = UriComponentsBuilder
        .fromUriString("{logic app url}")
        // Add query parameter
        .queryParam("api-version", {api-version})
        .queryParam("sp", {sp})
        .queryParam("sv", {sv})
        .queryParam("sig",{sig});

RequestDetailsDAO requestDetails = new RequestDetailsDAO(builder.build().toUri().toString(), HttpMethod.GET);
String response = httpClientService.execute(requestDetails, null, responseErrorHandler, String.class);
return response.toString();

HttpClientService.java

@Service
public class HttpClientService<T, V>  {
public RestTemplate restTemplate;

public HttpClientService(RestTemplateBuilder restTemplateBuilder) {
    this.restTemplate = restTemplateBuilder.setConnectTimeout(Duration.ofSeconds(5)).setReadTimeout(Duration.ofSeconds(5)).build();
}

public V execute(RequestDetailsDAO requestDetails, HttpEntity<T> entity, ResponseErrorHandler errorHandler,
                 Class<V> genericClass) {

    restTemplate.setErrorHandler(errorHandler);

    ResponseEntity<V> response = restTemplate.exchange(requestDetails.getUrl(), requestDetails.getRequestType(), entity, genericClass);
    return response.getBody();
}
}

请求详情DAO.java

public class RequestDetailsDAO {
private String url;
private HttpMethod requestType;

public RequestDetailsDAO(String url, HttpMethod requestTyp) {
    super();
    this.url = url;
    this.requestType = requestTyp;
}
public String getUrl() {
    return url;
}
public void setUrl(String url) {
    this.url = url;
}
public HttpMethod getRequestType() {
    return requestType;
}
public void setRequestType(HttpMethod requestType) {
    this.requestType = requestType;
}

@Override
public String toString() {
    return "RequestDetails [url=" + url + ", requestType=" + requestType + "]";
}
}

Answer 1

请尝试使用带有 RestTemplate 的简单独立代码并检查。 我提供了下面的小片段。

try {
    ResponseEntity<ResponseVO> response = restTemplate.exchange({uri of azure logic app trigger}, HttpMethod.GET, request, ResponseVO.class);
} catch (HttpStatusCodeException ex) {
    int statusCode = ex.getStatusCode().value();
    System.out.println("Status Code :"+statusCode);
    ResponseEntity<?> resEntity = ResponseEntity.status(ex.getRawStatusCode()).headers(ex.getResponseHeaders())
                .body(ex.getResponseBodyAsString())
}

这里的ResponseVO.class是要映射到object的Response，在这种情况下，您可以设置自己的class。 在这个 catch 块中，您可以找到异常详细信息。

Answer 2

因此，使用 RestTemplate 调用 LogicApp 似乎没有什么特别之处。 只是 RestTemplate 默认 URL 对参数中给出的 Uri 进行编码，而 DefaultHttpClient 没有 - 参考： apache vs resttemplate

在 LogicApp URL 的情况下，有这个“sp”参数已经是 URL 结束编码，当你从 LogicApp -“%2Ftriggers%2Fmanual%2Frun”复制它时，所以你需要解码并通过“/手动/触发” " 到 UriComponentsBuilder。 然后它就起作用了。

我的代码：

    RestTemplate restTemplate = new RestTemplate();
    HttpHeaders headers = new HttpHeaders();
    headers.setContentType(MediaType.APPLICATION_JSON);
    HttpEntity<Object> request = new HttpEntity<>(requestDto, headers);

    UriComponentsBuilder builder = UriComponentsBuilder
            .fromUriString("https://prod-123901.westeurope.logic.azure.com:443/workflows/<workflow_id>/triggers/manual/paths/invoke")
            .queryParam("api-version", "2016-10-01")
            .queryParam("sp", "/triggers/manual/run")
            .queryParam("sv", "1.0")
            .queryParam("sig", "<your_sig>");
    restTemplate.exchange(builder.build().toUri().toString(), HttpMethod.POST, request, Void.class);

更新：

requestDto 这里是您的自定义 dto object 进入 HTTP 主体，例如 object 的 ZA2F2ED4F8EBC2CBBDZC21:

public class RequestDto {
    private String id;
    private String name;

    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }
}