我无法使用amazon java SDK和IAm用户凭据从我的亚马逊弹性搜索群集中获取结果。现在的问题是,当PATH字符串等于“/”时,我能够正确获取结果但是当我尝试使用不同的路径进行例如“/ private-search”时,我得到403禁止错误。即使对于具有公共访问权限的路径,我也会为此IAm用户收到403禁用错误,但如果我删除“signer.sign(requestToSign,credentials);”在performSigningSteps方法中的行(仅适用于公共资源)。
我在AWS中的政策使此IAM用户可以访问我的弹性搜索服务中的所有内容。还有什么办法可以避免在源代码中对访问密钥和密钥进行硬编码?
private static final String SERVICE_NAME = "es";
private static final String REGION = "region-name";
private static final String HOST = "host-name";
private static final String ENDPOINT_ROOT = "http://" + HOST;
private static final String PATH = "/private-search";
private static final String ENDPOINT = ENDPOINT_ROOT + PATH;
private static String accessKey = "IAmUserAccesskey"
private static String secretKey = "IAmUserSecretkey"
public static void main(String[] args) {
// Generate the request
Request<?> request = generateRequest();
// Perform Signature Version 4 signing
performSigningSteps(request);
// Send the request to the server
sendRequest(request);
}
private static Request<?> generateRequest() {
Request<?> request = new DefaultRequest<Void>(SERVICE_NAME);
request.setContent(new ByteArrayInputStream("".getBytes()));
request.setEndpoint(URI.create(ENDPOINT));
request.setHttpMethod(HttpMethodName.GET);
return request;
}
private static void performSigningSteps(Request<?> requestToSign) {
AWS4Signer signer = new AWS4Signer();
signer.setServiceName(requestToSign.getServiceName());
signer.setRegionName(REGION);
AWSCredentials credentials = new BasicAWSCredentials(accessKey, secretKey);
signer.sign(requestToSign, credentials);
}
private static void sendRequest(Request<?> request) {
ExecutionContext context = new ExecutionContext();
ClientConfiguration clientConfiguration = new ClientConfiguration();
AmazonHttpClient client = new AmazonHttpClient(clientConfiguration);
MyHttpResponseHandler<Void> responseHandler = new MyHttpResponseHandler<Void>();
MyErrorHandler errorHandler = new MyErrorHandler();
Void response = client.execute(request, responseHandler, errorHandler, context);
}
public static class MyHttpResponseHandler<T> implements HttpResponseHandler<AmazonWebServiceResponse<T>> {
@Override
public AmazonWebServiceResponse<T> handle(com.amazonaws.http.HttpResponse response) throws Exception {
InputStream responseStream = response.getContent();
String responseString = convertStreamToString(responseStream);
System.out.println(responseString);
AmazonWebServiceResponse<T> awsResponse = new AmazonWebServiceResponse<T>();
return awsResponse;
}
@Override
public boolean needsConnectionLeftOpen() {
return false;
}
}
public static class MyErrorHandler implements HttpResponseHandler<AmazonServiceException> {
@Override
public AmazonServiceException handle(com.amazonaws.http.HttpResponse response) throws Exception {
System.out.println("In exception handler!");
AmazonServiceException ase = new AmazonServiceException("exception.");
ase.setStatusCode(response.getStatusCode());
ase.setErrorCode(response.getStatusText());
return ase;
}
@Override
public boolean needsConnectionLeftOpen() {
return false;
}
}
public static String convertStreamToString(InputStream is) throws IOException {
// To convert the InputStream to String we use the
// Reader.read(char[] buffer) method. We iterate until the
// Reader return -1 which means there's no more data to
// read. We use the StringWriter class to produce the string.
if (is != null) {
Writer writer = new StringWriter();
char[] buffer = new char[1024];
try {
Reader reader = new BufferedReader(new InputStreamReader(is, "UTF-8"));
int n;
while ((n = reader.read(buffer)) != -1) {
writer.write(buffer, 0, n);
}
}
finally {
is.close();
}
return writer.toString();
}
return "";
}