swift - 如何在 Swift 中使用 Amazon SES 发送电子邮件？

Question

我正在尝试使用 Amazon SES 发送电子邮件。运行此程序时我没有收到任何错误，但什么也没有发生。我试图找到文档，但亚马逊在这方面做得很糟糕。那么，有什么想法吗？编辑:我也在 Amazon SES 控制台中验证了我尝试发送的邮件。

这是我目前所得到的:

    let sns = AWSSES.defaultSES()

    var messageBody = AWSSESContent()
    var subject = AWSSESContent()
    var body = AWSSESBody()
    subject.data = "Subject"
    messageBody.data = "Message body"
    body.text = messageBody

    var message = AWSSESMessage()
    message.subject = subject
    message.body = body

    var destination = AWSSESDestination()
    destination.toAddresses = ["emailadress@gmail.com"]

    var send = AWSSESSendEmailRequest()
    send.source = "emailadress@gmail.com"
    send.destination = destination
    send.message = message


    sns.sendEmail(send).continueWithSuccessBlock {(task: BFTask!) -> BFTask! in
        NSLog("Sent mail - success")
        return nil
    }


}

Answer 1

所以，问题是我没有为 SES 服务使用正确的区域。工作代码:

let credentialProvider = AWSCognitoCredentialsProvider.credentialsWithRegionType(
            CognitoRegionType,
            identityPoolId: CognitoIdentityPoolId)
        let configuration = AWSServiceConfiguration(
            region: SESServiceRegionType,
            credentialsProvider: credentialProvider)
        var sns = AWSSES(configuration: configuration)

        var messageBody = AWSSESContent()
        var subject = AWSSESContent()
        var body = AWSSESBody()
        subject.data = "Subject"
        messageBody.data = "Message body"
        body.text = messageBody

        var theMessage = AWSSESMessage()
        theMessage.subject = subject
        theMessage.body = body

        var destination = AWSSESDestination()
        destination.toAddresses = ["emailadress@gmail.com"]

        var send = AWSSESSendEmailRequest()
        send.source = "emailadress@gmail.com"
        send.destination = destination
        send.message = theMessage
        send.returnPath = "emailadress@gmail.com"

        sns.sendEmail(send).continueWithBlock {(task: AnyObject!) -> BFTask! in

            return nil
        }