java - 将 Apache Log 时间转换为 Epoch 时间

Question

编辑:工作解决方案。原始问题如下。

private Timestamp extractTimestamp(Object timestamp) {
    try {
        return Timestamp.from(Instant.ofEpochMilli(Long.valueOf(String.valueOf(timestamp))));
    } catch(NumberFormatException e) {
        System.out.println("Not in Epoch format");
    }
    try {
        return Timestamp.from(Instant.parse(String.valueOf(timestamp)));
    } catch(DateTimeParseException e) {
        System.out.println("Not in UTC format");
    }
    try {
        return Timestamp.valueOf(String.valueOf(timestamp));
    } catch(IllegalArgumentException e) {
        System.out.println("Not in SQL format");
    }
    try {               
        SimpleDateFormat formatter = new SimpleDateFormat("dd/MMM/yyyy:hh:mm:ss Z");
        Date date = formatter.parse(String.valueOf(timestamp));
        return Timestamp.from(Instant.ofEpochMilli(date.getTime()));
    } catch(ParseException e) {
        System.out.println("Not in Apache Log format");
    }
    // Return current time if none found
    return Timestamp.from(Instant.now());
}

<小时/>

我正在尝试从 Apache 访问日志中解析时间戳并将其转换为 Epoch 时间戳或 SQL 时间戳。我已经有代码可以从纪元转换为其他格式的 SQL 时间戳，因此我主要关心的是获取纪元格式或任何其他易于转换的格式。我目前正在使用 Grok 模式，但我正在寻找一种更有效的提取时间的方法。

下面是我正在提取的日志和时间戳以及我当前的代码的示例:

127.0.0.1 127.0.0.1 - - [04/Nov/2016:08:00:02 -0400] "GET/loc/ation"200 163 "-""-"26 163 37526

04/11/2016:08:00:02 -0400

private Timestamp extractTimestamp(Object timestamp) {
  try {
    return Timestamp.from(Instant.ofEpochMilli(Long.valueOf(String.valueOf(timestamp))));
  } catch(NumberFormatException e) {
    System.out.println("Not in Epoch format");
  }
  try {
    return Timestamp.from(Instant.parse(String.valueOf(timestamp)));
  } catch(DateTimeParseException e) {
    System.out.println("Not in UTC format");
  }
  try {
    return Timestamp.valueOf(String.valueOf(timestamp));
  } catch(IllegalArgumentException e) {
    System.out.println("Not in SQL Time format");
  }
  try {
    // Sample timestamp: 04/Nov/2016:08:00:02 -0400
    String apacheLogExpression = "%{NUMBER:day}/%{WORD:month}/%{NUMBER:year}:%{NUMBER:hour}:%{NUMBER:minute}:%{NUMBER:second}\\s%{GREEDYDATA:offset}";
    Grok compiledPattern = dictionary.compileExpression(apacheLogExpression);
    Map<String, String> values = compiledPattern.extractNamedGroups(String.valueOf(timestamp));
    System.out.println(values);
  } catch(Exception e) {
    System.out.println("Not in Apache Log format");
    e.printStackTrace();
  }
  // Return current time if none found
  return Timestamp.from(Instant.now());
}

预先感谢您的帮助!

Answer 1

    String logTime = "04/Nov/2016:08:00:02 -0400";
    SimpleDateFormat formatter = new SimpleDateFormat("dd/MMM/yyyy:hh:mm:ss Z");
    Date date = formatter.parse(logTime);
    System.out.println(date);

将打印 Fri Nov 04 14:00:02 EET 2016 并基本上获取 java.util.Date 对象