如何使用 Vapor Websockets 向客户端发送字典 [英] How Do I Send a Dictionary to a Client using Vapor Websockets

问题描述

我已经在这里继续这个问题，因为重点已经改变，但是是相关的.

I've continued this question here, because the focus has changed, but is related.

Vapor 正在服务器和 iOS 客户端之间进行通信.只是设置代码，用于学习目的.

Vapor is communicating between a server and and iOS client. Just setup code, for learning purposes.

现在，我想使用 JSON 通过已建立的连接发送值字典.我陷入了无法解释的演示代码逻辑中，但在我的蒸气路线中，这就是我所处的位置:

Now, I want to send a dictionary of values via the established connection using JSON. I'm getting caught up in the unexplained logic of demo code, but here's where I am, in my vapor routes:

    app.webSocket("respond") { req, ws in
        ws.onText { ws, text in
            print(text)
        
            let msgDict = "{'name' = 'bobby'}"
            let encoder = JSONEncoder()
            let data = try encoder.encode(msgDict)
                
            ws.send(data)
        }
    }

这将无法编译: Invalid conversion from throwing function of type '(WebSocket, String) throws ->()' 到非抛出函数类型 '(WebSocket, String) ->()'

虽然我通常理解这个错误，并且取决于我如何玩这个错误.

while I generally understand this error, and dependent on how I play with this it varies.

通常我要寻找的是一个简单的模式来获取内部字典值，将它们编码为 JSON，然后发送它们.什么是我没有做的，或者我做的不对?

What generally I'm looking for is a simple pattern to take internal dictionary values, encode them into JSON, and send them. What am I not doing, or not doing right?

推荐答案

我发现该代码存在两个问题.

I see two problems with that code.

第一个，由编译错误解释，它实际上告诉它不会处理抛出的任何错误.当您执行 encoder.encode(msgDict) 时，此代码可能会抛出 Error，并且您不会在任何地方处理此可能的错误.如果您处理了那个可能的错误，您在该闭包中编写的代码将具有预期的类型.您有几种选择:

The first one, explained by the compiling error, it's actually telling that it will not handle any errors thrown. When you do encoder.encode(msgDict), this code can throw an Error, and you're not handling this possible error anywhere. If you handle that possible error, code you wrote in that closure will have the expected type. You have a few options to do so:

• 将代码包裹在 do-catch 块中

do {
    let data = try encoder.encode(msgDict)
    ws.send(data)
} catch let error {
    // Handle error
}

• 使用try?，意味着如果发生错误，您不处理错误而是得到nil.
• Use try?, means you do not handle the error but get nil as a result, if an error occurred.

if let data = try? encoder.encode(msgDict) {
    ws.send(data)
}

• 使用 try!，意味着您强制解包结果(不建议，但您可以尝试，如果您 100% 确定)
• Use try!, means you force-unwrap the result (not advisable, but you can try, if you're 100% sure)

let data = try! encoder.encode(msgDict)
ws.send(data)

第二个问题是您如何编写该响应 - "{'name' = 'bobby'}".这是一个无效的 JSON，您应该使用双引号:

The second problem is how you're writing that response - "{'name' = 'bobby'}". This is an invalid JSON, you should use double quotes instead:

let msgDict = "{\"name\" = \"bobby\"}"

你也可以使用Dictionary，只要内容是Encodable类型:

You can also use Dictionary, as long as the content is of type Encodable:

let msgDict = ["name": "bobby"]

您还可以使用 JSONEncoder 对符合 Encodable 的类的任何实例进行编码.

You can also use JSONEncoder to encode any instances of classes that conform to Encodable.

所以整件事，我会这样写:

So the whole thing, I'd write like this:

app.webSocket("respond") { req, ws in
    ws.onText { ws, text in
        print(text)
    
        let msgDict = ["name": "bobby"]
        let encoder = JSONEncoder()
        do {
            let data = try encoder.encode(msgDict)
            ws.send(data)
        }
        catch let error {
            print("An error occurred: \(error)")
        }
    }
}

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