Question

我在Spring @Controller上有以下方法：

@MessageMapping("/comment/{id}")
@SendTo("/topic/conversation/{id}")
public OperationResult<Comment> comment(Comment comment) throws Exception {

    //call @Service to add comment
    OperationResult<Comment> operationResult = commentService.comment(comment);

    return operationResult;}

即使操作不成功，这也会向所有订阅者发送更新（operationResult.success == false）。

NB： OperationResult有一个名为“success”的布尔字段，用于指示操作是否成功。

问题： 我想知道如果操作成功，我怎么才能向订阅者发送更新（不抛出异常）并确保发送评论的客户端始终获得 operationResult

Answer 1

我设法找到了一种只根据条件发送更新的方法

@Autowired
private SimpMessagingTemplate simpMessagingTemplate;

@MessageMapping("/comment/{id}" )
public OperationResult<Comment> comment(Comment comment) throws Exception {

    OperationResult<Comment> operationResult = new OperationResult<>();
    conversationService.comment(operationResult, comment);

    if (operationResult.isSuccess()) {
        simpMessagingTemplate.convertAndSend("/topic/conversation/"+operationResult.getReturnedObject().getConversation().getId(), operationResult);
    }
    return operationResult;
}

然而，这不允许我将 operationResult 返回给客户端。

<强>更新

为了让客户端获得 operationResult ，我将@MessageMapping替换为@RequestMapping。因此，客户必须进行正常的ajax调用才能提交评论。

@Autowired
private SimpMessagingTemplate simpMessagingTemplate;

@RequestMapping("/comment" )
public OperationResult<Comment> comment(@RequestBody Comment comment) throws Exception {

    OperationResult<Comment> operationResult = new OperationResult<>();
    conversationService.comment(operationResult, comment);

    if (operationResult.isSuccess()) {
        simpMessagingTemplate.convertAndSend("/topic/conversation/"+operationResult.getReturnedObject().getConversation().getId(), operationResult);
    }
    return operationResult;
}

Spring-Websocket：满足条件时向订阅者发送更新

1 个答案: