(Swift) twilio post 请求使用 alamofire 设置属性

(Swift) twilio post request to set attributes using alamofire

我正在尝试使用 Twilio 的 REST Api 和 Alamofire 在创建频道时将某些属性设置为频道 (https://www.twilio.com/docs/api/ip-messaging/rest/channels#action-create)

let parameters : [String : AnyObject] = [
    "FriendlyName": "foo",
    "UniqueName": "bar",
    "Attributes": [
        "foo": "bar",
        "bar": "foo"
    ],
    "Type": "private"
]

Alamofire.request(.POST, "https://ip-messaging.twilio.com/v1/Services/\(instanceSID)/Channels", parameters: parameters)
    .authenticate(user: user, password: password)
    .responseJSON { response in
        let response = String(response.result.value)
        print(response)
}

使用该代码,我收到的回复是创建了一个带有 FriendlyName foo 和 UniqueName bar 的频道,但该频道没有设置属性。

查看 Alamofire github (https://github.com/Alamofire/Alamofire) 我发现有一种方法可以发送带有 JSON 编码参数的 POST 请求。所以我尝试了这个:

let parameters : [String : AnyObject] = [
    "FriendlyName": "foo",
    "UniqueName": "bar15",
    "Attributes": [
        "foo": "bar",
        "bar": "foo"
    ],
    "Type": "private"
]

Alamofire.request(.POST, "https://ip-messaging.twilio.com/v1/Services/\(instanceSID)/Channels", parameters: parameters, encoding: .JSON)
    .authenticate(user: user, password: password)
    .responseJSON { response in
        let response = String(response.result.value)
        print(response)
}

向请求添加 "encoding: .JSON" 时,响应显示不仅未设置属性,而且 FriendlyName 和 UniqueName 也为零,这与之前使用 URL 正确设置它们不同 -编码参数。

我是否在 'parameters' 中设置了错误的属性? Twilio 的文档说属性是 "An optional metadata field you can use to store any data you wish. No processing or validation is done on this field."

帮助将不胜感激:)

我找到了问题的答案。原来我的属性字段格式不正确。

这是对我有用的代码:

let parameters : [String : AnyObject] = [
    "FriendlyName": "foo",
    "UniqueName": "bar",
    "Attributes": "{\"key\":\"value\",\"foo\":\"bar\"}",
    "Type": "private"
]

Alamofire.request(.POST, "https://ip-messaging.twilio.com/v1/Services/\(instanceSID)/Channels/", parameters: parameters)
    .authenticate(user: user, password: password)
    .responseJSON { response in
        let response = String(response.result.value)
        print(response)
        debugPrint(response)
}

希望这对其他人有帮助:)