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我需要通过 HTTP 获取一个 excel 电子表格,然后将其作为附件发送到我的 Java Spring 服务器上的电子邮件中。

我发现的问题是 MultiPartEmail.attach() 只需要一个 java.net.URL 实例,我无法弄清楚如何确保请求在其标头中有特定的 Cookie 以进行身份​​验证。

url = new URL(urlString);
email.attach(url, "test.xls", "File");
email.send();

我试图手动请求并创建一个工作簿,但后来我很难将工作簿本身附加到 MultiPartEmail。

HttpClient client = new HttpClient();
GetMethod method = new GetMethod(queryString);

method.setRequestHeader("Cookie", cookie);
client.executeMethod(method);

InputStream stream = method.getResponseBodyAsStream();

Workbook workbook = Workbook.getWorkbook(stream);
email.attach(workbook, "report.xls", "forecasting report");

我需要一些方法来解决这些限制。

在此先感谢您的时间。

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1 回答 1

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由于您使用的是 Spring,因此您可以使用其内置的电子邮件支持来发送电子邮件。因此,您如何从磁盘或其他地方检索文件并不重要。您可以使用 MimeMessageHelper 发送带有附件的电子邮件,并为邮件主机上的帐户指定用户名和密码以进行身份​​验证,如下所示:

public class EmailNotifier {
private JavaMailSenderImpl mailSender;

public void setMailSender(JavaMailSenderImpl mailSender) {
    this.mailSender = mailSender;
}

public void sendMail(InternetAddress fromAddress, InternetAddress toAddress, String subject, String msg) {

    MimeMessage message = mailSender.createMimeMessage();

    try {
        // use the true flag to indicate you need a multipart message
        MimeMessageHelper helper = new MimeMessageHelper(message, true);

        helper.setFrom(fromAddress);
        helper.setTo(toAddress);
        helper.setSubject(subject);
        helper.setText(msg);
        // let's attach the infamous windows Sample file (this time copied to c:/)
        FileSystemResource file = new FileSystemResource(new File("c:/test.xls"));
        helper.addAttachment("test.xls", file);
        mailSender.send(message);
    } catch (MessagingException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }           
}
}

并在您的 bean 配置文件中配置一个 JavaMailSenderImpl bean。这是通过 Gmail 发送电子邮件。

豆类.xml

<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans.xsd">

    <bean id="mailSender" class="org.springframework.mail.javamail.JavaMailSenderImpl">
        <property name="host" value="smtp.gmail.com" />
        <property name="port" value="587" />
        <property name="username" value="adminxxxx" />
        <property name="password" value="password" />

        <property name="javaMailProperties">
           <props>
                  <prop key="mail.smtp.auth">true</prop>
                  <prop key="mail.smtp.starttls.enable">true</prop>
               </props>
        </property>
    </bean>

    <bean id="emailNotifier" class="com.examples.EmailNotifier">
        <property name="mailSender" ref="mailSender" />
    </bean>

</beans>

这只是为了测试:

public class MailApp {
  public static void main( String[] args )
    {
        ApplicationContext context = 
             new ClassPathXmlApplicationContext("beans.xml");

        EmailNotifier emailNotifier = (EmailNotifier) context.getBean("emailNotifier");
        try {
            emailNotifier.sendMail(new InternetAddress("adminxxxx@gmail.com"),
                    new InternetAddress("recipientxxxx@gmail.com"),
                   "Email Alert!", 
                   "Hello User, This is a test email \n No response required.");
        } catch (AddressException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }    
    }

弹簧文档

于 2012-08-01T05:05:11.140 回答