英文:
Why does my event run twice when data only fed once?
问题
所以，我已经修整了我的Go代码，以便更好地理解，因为我正在使用库来实现这个功能。
基本上，我正在尝试接收一个JSON文本，将其修整成特定的格式，并将其传递给createTicket函数。
一切都正常工作，我的createTicket函数使用replyText中的正确主题字段创建了一个ticket，但是每当我将一个JSON输入到我的应用程序中时，createTicket函数运行了2-3次，并且我的调试打印消息打印了几次。
在后端，当我只在测试期间一次性输入一个JSON时，createTicket将在我的系统上创建一个ticket，并且会出现2-3个新的ticket。
func handleMessage(bot *lark.Bot, msg *lark.EventV2MessageReceived, client *zendesk.Client) {
query := msg.Message.Content
//fmt.Println(query)
var replyText ReplyText
err := json.Unmarshal([]byte(query), &replyText)
if err != nil {
log.Fatal(err)
}
fmt.Println(replyText.Text)
client.CreateTicket(context.Background(), zendesk.Ticket{
Subject: replyText.Text,
Priority: "P2",
Comment: zendesk.TicketComment{
Body: "testBody",
},
})
//print("DEBUG: MSG HANDLED AND TICKET CREATED")
}
调用handleMsg()函数的代码是通过以下API调用在我的服务器上监听的。基本上只是一个事件处理程序，所以当我发送一条消息到我的应用程序时，它将进行必要的检查，并将接收到的消息传递给函数，如下所示。
r.POST("/", func(c *gin.Context) {
if evt, ok := middleware.GetEvent(c); ok {
if evt.Header.EventType == lark.EventTypeMessageReceived {
if msg, err := evt.GetMessageReceived(); err == nil {
fmt.Println(msg.Message.Content)
handleMessage(bot, msg, client)
c.JSON(http.StatusOK, gin.H{"status": "ok"})
}
}
}
})
英文:
So i have trimmed my Go code below for better understanding as I am using libraries to implement this.
Basically what I am trying to do is receive a JSON text, trim it into a certain format and pass it over to createTicket function.
Everything works fine and my createTicket function creates it with the correct subject field from replyText, however createTicket runs almost 2-3x everytime I feed a JSON into my application and my debug print message prints few times.
On the backend createTicket will create a ticket on my system and 2-3 new tickets appear when I have only fed the system a JSON only once during my testing.
func handleMessage(bot *lark.Bot, msg *lark.EventV2MessageReceived, client *zendesk.Client) {
query := msg.Message.Content
//fmt.Println(query)
var replyText ReplyText
err := json.Unmarshal([]byte(query), &replyText)
if err != nil {
log.Fatal(err)
}
fmt.Println(replyText.Text)
client.CreateTicket(context.Background(), zendesk.Ticket{
Subject: replyText.Text,
Priority: "P2",
Comment: zendesk.TicketComment{
Body: "testBody",
},
})
//print("DEBUG: MSG HANDLED AND TICKET CREATED")
}
The code that calls handleMsg() function is through an API call below listening on my server. Basically just an event handler, So when I send a message to my application below it will do the necessary checks and I will pass the message received to the function as below.
r.POST("/", func(c *gin.Context) {
if evt, ok := middleware.GetEvent(c); ok {
if evt.Header.EventType == lark.EventTypeMessageReceived {
if msg, err := evt.GetMessageReceived(); err == nil {
fmt.Println(msg.Message.Content)
handleMessage(bot, msg, client)
c.JSON(http.StatusOK, gin.H{"status": "ok"})
}
}
}
})
答案1
得分: 0
我在 handleMessage(bot, msg, client) 中遗漏了 go。
应该是 go handleMessage(bot, msg, client) 来调用该例程。
这样修复了我的问题。
英文:
I was missing go in handleMessage(bot, msg, client)
It should be go handleMessage(bot, msg, client) to call the routine once.
This fixed my issue.
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论