如何从电子邮件中删除引用的文本,仅显示新文本

Question

我正在解析电子邮件.当我看到对电子邮件的回复时,我想删除引用的文本,以便我可以将文本附加到上一封电子邮件中(即使是回复).

通常情况下,您会看到:

第一封电子邮件(会话开始)

This is the first email

第2封电子邮件(回复第一封)

This is the second email

Tim said:
This is the first email

这个输出只是"这是第二封电子邮件".虽然不同的电子邮件客户端引用文本的方式不同,但如果只是为了获得大部分新的电子邮件文本,那也是可以接受的.

Answer 1

我使用以下正则表达式匹配引用文本的引导(最后一个是重要的):

  /** general spacers for time and date */
  private static final String spacers = "[\\s,/\\.\\-]";

  /** matches times */
  private static final String timePattern  = "(?:[0-2])?[0-9]:[0-5][0-9](?::[0-5][0-9])?(?:(?:\\s)?[AP]M)?";

  /** matches day of the week */
  private static final String dayPattern   = "(?:(?:Mon(?:day)?)|(?:Tue(?:sday)?)|(?:Wed(?:nesday)?)|(?:Thu(?:rsday)?)|(?:Fri(?:day)?)|(?:Sat(?:urday)?)|(?:Sun(?:day)?))";

  /** matches day of the month (number and st, nd, rd, th) */
  private static final String dayOfMonthPattern = "[0-3]?[0-9]" + spacers + "*(?:(?:th)|(?:st)|(?:nd)|(?:rd))?";

  /** matches months (numeric and text) */
  private static final String monthPattern = "(?:(?:Jan(?:uary)?)|(?:Feb(?:uary)?)|(?:Mar(?:ch)?)|(?:Apr(?:il)?)|(?:May)|(?:Jun(?:e)?)|(?:Jul(?:y)?)" +
                                              "|(?:Aug(?:ust)?)|(?:Sep(?:tember)?)|(?:Oct(?:ober)?)|(?:Nov(?:ember)?)|(?:Dec(?:ember)?)|(?:[0-1]?[0-9]))";

  /** matches years (only 1000's and 2000's, because we are matching emails) */
  private static final String yearPattern  = "(?:[1-2]?[0-9])[0-9][0-9]";

  /** matches a full date */
  private static final String datePattern     = "(?:" + dayPattern + spacers + "+)?(?:(?:" + dayOfMonthPattern + spacers + "+" + monthPattern + ")|" +
                                                "(?:" + monthPattern + spacers + "+" + dayOfMonthPattern + "))" +
                                                 spacers + "+" + yearPattern;

  /** matches a date and time combo (in either order) */
  private static final String dateTimePattern = "(?:" + datePattern + "[\\s,]*(?:(?:at)|(?:@))?\\s*" + timePattern + ")|" +
                                                "(?:" + timePattern + "[\\s,]*(?:on)?\\s*"+ datePattern + ")";

  /** matches a leading line such as
   * ----Original Message----
   * or simply
   * ------------------------
   */
  private static final String leadInLine    = "-+\\s*(?:Original(?:\\sMessage)?)?\\s*-+\n";

  /** matches a header line indicating the date */
  private static final String dateLine    = "(?:(?:date)|(?:sent)|(?:time)):\\s*"+ dateTimePattern + ".*\n";

  /** matches a subject or address line */
  private static final String subjectOrAddressLine    = "((?:from)|(?:subject)|(?:b?cc)|(?:to))|:.*\n";

  /** matches gmail style quoted text beginning, i.e.
   * On Mon Jun 7, 2010 at 8:50 PM, Simon wrote:
   */
  private static final String gmailQuotedTextBeginning = "(On\\s+" + dateTimePattern + ".*wrote:\n)";


  /** matches the start of a quoted section of an email */
  private static final Pattern QUOTED_TEXT_BEGINNING = Pattern.compile("(?i)(?:(?:" + leadInLine + ")?" +
                                                                        "(?:(?:" +subjectOrAddressLine + ")|(?:" + dateLine + ")){2,6})|(?:" +
                                                                        gmailQuotedTextBeginning + ")"
                                                                      );

我知道在某些方面这是过度的(可能会很慢!)但它的效果非常好.如果您发现任何与此不符的内容,请告诉我,以便我可以改进!

Answer 2

查看Google专利:http://www.google.com/patents/US7222299

总之,它们对文本的部分进行散列(可能类似于句子),然后在先前的消息中查找与散列的匹配.速度非常快,他们也可能将此作为线程算法的输入.真是个好主意!

Answer 3

当以前的电子邮件存储在磁盘上或可用时，您可以检查由特定收件人发送的所有邮件，以确定哪些是响应文本。

您还可以尝试通过检查最后几行的第一个字符来确定引号字符。通常，最后几行总是以相同的字符开头。

当最后两行以不同的字符开头时，您可以尝试第一行，因为有时答案会附加在文本末尾。

如果检测到这些字符，则可以删除以此字符开头的最后几行，直到检测到空行或以另一个字符开头的行。

未经测试，更像是伪代码

    String[] lines;

    // Check the size of the array first, length > 2
    char startingChar = lines[lines.length - 1].charAt(0);
    int foundCounter = 0;
    for (int i = lines.length - 2; i >=0; --i) {
        String line = lines[i];

        // Check line size > 0
        if(startingChar == line.charAt(0)){
            ++foundCounter;
        }
    }

    final int YOUR_DECISION = 2; // You can decide
    if(foundCounter > YOUR_DECISION){
        deleteLastLinesHere(startingChar, foundCounter);
    }