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我有access_logs，我500MB给样品

10.223.157.186 - - [15/Jul/2009:14:58:59 -0700] "GET / HTTP/1.1" 403 15779
10.223.157.186 - - [15/Jul/2009:14:58:59 -0700] "GET /favicon.ico HTTP/1.1" 404 5397
10.216.113.172 - - [29/Apr/2010:07:19:48 -0700] "GET / HTTP/1.1" 200 68831

如何从时间戳中提取月份？

预期输出：

year   month    day    event occurrence

2009   jul      15     GET /favicon.ico HTTP/1.1

2009   apr      29     GET / HTTP/1.1

我试过这个

add jar /usr/lib/hive/lib/hive-contrib-0.7.1-cdh3u2.jar;

create table log(ip string, gt string, gt1 string, timestamp string, id1 string, s1 string, s2 string) row format serde 'org.apache.hadoop.hive.contrib.serde2.RegexSerDe'                          
with serdeproperties('input.regex'= '^(\\S+) (\\S+) (\\S+) \\[([[\\w/]+:(\\d{2}:\\d{2}):\\d{2}\\s[+\\-]\\d{4}:/]+\\s[+\\-]\\d{4})\\] "(.+?)" (\\S+) (\\S+)')location '/path';

如果我理解正确，字符串函数在这种情况下将不起作用。我是regex&的新手hive。

帮帮我..提前谢谢

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1 回答 1

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我不熟悉 hadoop/hive，但就正则表达式而言，如果我使用的是 ruby​​：

log_file = %Q[
  10.223.157.186 - - [15/Jul/2009:14:58:59 -0700] "GET / HTTP/1.1" 403 15779
  10.223.157.186 - - [15/Jul/2009:14:58:59 -0700] "GET /favicon.ico HTTP/1.1" 404 5397
  10.216.113.172 - - [29/Apr/2010:07:19:48 -0700] "GET / HTTP/1.1" 200 68831
]

converted_lines = log_file.split("\n").map do |line|
  regex = /^.*? - - \[(\d+)\/(\w+)\/(\d{4}).*?\] (.*)/
  matches = regex.match(line)
  output = [
    [:year, matches[3]],
    [:month, matches[2]],
    [:day, matches[1]],
    [:event_occurrence, matches[4]],
  ]
end

希望有帮助。

于 2013-10-30T10:08:43.557 回答