Return 来自 JAX-RS Rest 服务的异常 JSON
Return exception from JAX-RS Rest Service as JSON
是否有可能以某种方式将休息服务抛出的异常作为 JSON 返回?我有一个 JAX-RS Rest 服务,我想在其中实现这一点。当我现在抛出它时,它被映射到一个 HTML 响应,这不是我想要的。据我了解,ExceptionMapper 也会将其映射到 HTML?是否有任何其他替代方法或库允许以 JSON 格式返回异常?
捕获异常,然后以标准化格式构建响应对象,例如
error: {
code: 'XXX',
status: HTTPStatus,
message: 'my error message'
}
并将其作为具有错误状态的响应发送(来自 Response.Status,通常为 4xx 或 5xx)
您可以创建自定义异常,需要 JSON 请求和响应
@POST
@Path("/betRequest")
@Consumes({ "application/json", "application/x-www-form-urlencoded" })
@Produces({ "application/json", "application/x-www-form-urlencoded" })
public Response getBetRequest(String betRequestParams, @Context HttpServletRequest request)
{
BetResponseDetails betResponseDetails = new BetResponseDetails();
try{
//you code here
}
catch (JSONException ex)
{
ex.printStackTrace();
betResponseDetails.setResponseCode("9002");//your custom error code
betResponseDetails.setResponseStatus("Bad Request");//custom status
betResponseDetails.setResponseMessage("The request body contained invalid JSON");//custom error massage
return Response.status(200).entity(betResponseDetails).build();
}
}
创建一个 POJO BetResponseDetails
public class BetResponseDetails {
private String ResponseStatus;
private String ResponseCode;
private String ResponseMessage;
// getter/setter
.......
}
获取状态和数据结构中的响应数据,如果状态错误,则显示正确的消息。
你可以这样试试
{
"status": "error",
"data": {
"message": "information of error message"
}
}
它将响应为 JSON。
@Provider
@Singleton
public class ExceptionMapperProvider implements ExceptionMapper<Exception>
{
@Override
public Response toResponse(final Exception exception)
{
return Response.status(HttpStatusCodes.STATUS_CODE_SERVER_ERROR).entity(new BasicResponse(InternalStatus.UNHANDLED_EXCEPTION, exception.getMessage())).type(MediaType.APPLICATION_JSON).build();
}
}
@XmlRootElement
public class BasicResponse {
public String internalStatus;
public String message;
public BasicResponse() {}
public BasicResponse(String internalStatus, String message){
this.internalStatus = internalStatus;
this.message = message;
}
}
是否有可能以某种方式将休息服务抛出的异常作为 JSON 返回?我有一个 JAX-RS Rest 服务,我想在其中实现这一点。当我现在抛出它时,它被映射到一个 HTML 响应,这不是我想要的。据我了解,ExceptionMapper 也会将其映射到 HTML?是否有任何其他替代方法或库允许以 JSON 格式返回异常?
捕获异常,然后以标准化格式构建响应对象,例如
error: {
code: 'XXX',
status: HTTPStatus,
message: 'my error message'
}
并将其作为具有错误状态的响应发送(来自 Response.Status,通常为 4xx 或 5xx)
您可以创建自定义异常,需要 JSON 请求和响应
@POST
@Path("/betRequest")
@Consumes({ "application/json", "application/x-www-form-urlencoded" })
@Produces({ "application/json", "application/x-www-form-urlencoded" })
public Response getBetRequest(String betRequestParams, @Context HttpServletRequest request)
{
BetResponseDetails betResponseDetails = new BetResponseDetails();
try{
//you code here
}
catch (JSONException ex)
{
ex.printStackTrace();
betResponseDetails.setResponseCode("9002");//your custom error code
betResponseDetails.setResponseStatus("Bad Request");//custom status
betResponseDetails.setResponseMessage("The request body contained invalid JSON");//custom error massage
return Response.status(200).entity(betResponseDetails).build();
}
}
创建一个 POJO BetResponseDetails
public class BetResponseDetails {
private String ResponseStatus;
private String ResponseCode;
private String ResponseMessage;
// getter/setter
.......
}
获取状态和数据结构中的响应数据,如果状态错误,则显示正确的消息。 你可以这样试试
{
"status": "error",
"data": {
"message": "information of error message"
}
}
它将响应为 JSON。
@Provider
@Singleton
public class ExceptionMapperProvider implements ExceptionMapper<Exception>
{
@Override
public Response toResponse(final Exception exception)
{
return Response.status(HttpStatusCodes.STATUS_CODE_SERVER_ERROR).entity(new BasicResponse(InternalStatus.UNHANDLED_EXCEPTION, exception.getMessage())).type(MediaType.APPLICATION_JSON).build();
}
}
@XmlRootElement
public class BasicResponse {
public String internalStatus;
public String message;
public BasicResponse() {}
public BasicResponse(String internalStatus, String message){
this.internalStatus = internalStatus;
this.message = message;
}
}