Question

当我使用aws ses发送电子邮件异常时，显示我

com.amazonaws.AmazonServiceException：无效日期星期一，2015年12月14日02:08:56 +00：00。它必须在其中一个 HTTP RFC 2616第3.3.1节规定的格式（服务： AmazonSimpleEmailService;状态代码：400;错误代码： InvalidParameterValue;请求ID： e2716096-a207-11e5-9615-8135b4d7f5f9）

以下是我的代码：

public class SESEmailUtil {

    private final String accesskey = "XXXXXXXXX";
    private final String secretkey = "XXXXXXXXXXXXXXXX";
    private String REGION = "us-east-1";
    private Region region;

    private static AWSCredentials credentials;
    private static AmazonSimpleEmailServiceClient sesClient;

    private static SESEmailUtil sesEmailUtil = null;

    private SESEmailUtil() { 
        init(accesskey, secretkey);
    };

    public void init(String accesskey, String secretkey) {
        credentials = new BasicAWSCredentials(accesskey, secretkey);
        sesClient = new AmazonSimpleEmailServiceClient(credentials);
        region = Region.getRegion(Regions.fromName(REGION));
        sesClient.setRegion(region);
    }

    public static SESEmailUtil getInstance() {
        if (sesEmailUtil == null) {
            synchronized (SESEmailUtil.class) {
                return new SESEmailUtil();
            }
        } else {
            return sesEmailUtil;
        }
    }

    public void sendEmail(String sender, LinkedList<String> recipients,
            String subject, String body) {
        Destination destination = new Destination(recipients);
        try {
            Content subjectContent = new Content(subject);
            Content bodyContent = new Content(body);
            Body msgBody = new Body(bodyContent);
            Message msg = new Message(subjectContent, msgBody);

            SendEmailRequest request = new SendEmailRequest(sender,
                    destination, msg);

            SendEmailResult result = sesClient.sendEmail(request);

            System.out.println(result + "Email sent");
        } catch (Exception e) {
            e.printStackTrace();
            System.out
                    .println("Exception from EmailSender.java. Email not send");
        }
    }
}


public class TestSend {
    private static String sender = "";
    private static LinkedList<String> recipients = new LinkedList<String>();
    static final String BODY = "This email was sent through Amazon SES by using the AWS SDK for Java.";
    static final String SUBJECT = "Amazon SES test (AWS SDK for Java)";
    public static void main(String args[]) {
        SESEmailUtil sendUtil = SESEmailUtil.getInstance();
        String receive = "qinwanghao@XXXX.com.cn";
        recipients.add(receive);
        sendUtil.sendEmail(sender, recipients, SUBJECT, BODY);
    }
}

代码基于aws提供的示例。日期2015年12月14日星期一02:08:56 +00：00无效，但我在哪里可以修改格式？希望有人能帮助我.THK。

Answer 1

Joda-time版本可能会出现问题。如果出现此错误，您应该使用maven（mvn dependency：tree）检查依赖树。寻找任何与aws-sdk提供的版本冲突的joda-time版本。修复：将joda-time的冲突版本添加到pom.xml中的排除项（或等效项）。

aws ses java无效的日期

1 个答案: