Question

我有一个这样的电子邮件日志文件：

2013-09-11 12:02:08  INFO: ------------------------------
2013-09-11 12:02:08  INFO: Javamail session sending email
2013-09-11 12:02:08  INFO: Session properties: 
2013-09-11 12:02:08  INFO:    com.hof.email.starttime=20130911120208
2013-09-11 12:02:08  INFO:    mail.smtp.auth=true
2013-09-11 12:02:08  INFO:    mail.smtp.connectiontimeout=60000
2013-09-11 12:02:08  INFO:    mail.smtp.host=mailserver
2013-09-11 12:02:08  INFO:    mail.smtp.port=25
2013-09-11 12:02:08  INFO:    mail.smtp.timeout=60000
2013-09-11 12:02:08  INFO:    mail.transport.protocol=smtp
2013-09-11 12:02:08  INFO: From: Support
2013-09-11 12:02:08  INFO: To: Customer
2013-09-11 12:02:08  INFO: Subject: Your Report Data
2013-09-11 12:02:08  INFO: Message ID: <id>
2013-09-11 12:02:09  INFO: Email sent successfully
2013-09-11 12:02:09  INFO: Javamail session ended
2013-09-11 12:02:09  INFO: ------------------------------

如果电子邮件主题与特定字符串匹配，我需要做的是打印整个记录。

也就是说，我认为我想做的是，当Subject = 'Your Report Data'时，然后打印'------------------------------'的第n次出现和'------------------------------'的第1次出现之间的所有行。来自主题匹配。

Answer 1

如果行之间的部分始终相同，则可以将grep与-A和-B一起使用。

Answer 2

这仅适用于gawk

awk '/Subject: Your Report Data/{printf "%s%s\n",$0,RT}' RS="------------------------------" file

编辑：更复杂的版本，打印正确的部分

cat file
2013-09-11 12:02:08  INFO: ------------------------------
2013-09-11 12:02:08  INFO: Javamail session sending email
2013-09-11 12:02:08  INFO: Session properties:
2013-09-11 12:02:08  INFO:    com.hof.email.starttime=20130911120208
2013-09-11 12:02:08  INFO:    mail.smtp.auth=true
2013-09-11 12:02:08  INFO:    mail.smtp.connectiontimeout=60000
2013-09-11 12:02:08  INFO:    mail.smtp.host=mailserver
2013-09-11 12:02:08  INFO:    mail.smtp.port=25
2013-09-11 12:02:08  INFO:    mail.smtp.timeout=60000
2013-09-11 12:02:08  INFO:    mail.transport.protocol=smtp
2013-09-11 12:02:08  INFO: From: Support
2013-09-11 12:02:08  INFO: To: Customer
2013-09-11 12:02:08  INFO: Subject: Your Report Data
2013-09-11 12:02:08  INFO: Message ID: <id>
2013-09-11 12:02:09  INFO: Email sent successfully
2013-09-11 12:02:09  INFO: Javamail session ended
2013-09-11 12:02:09  INFO: ------------------------------
2013-09-11 12:02:08  INFO: Javamail session sending email
2013-09-11 12:02:08  INFO: Session properties:
2013-09-11 12:02:08  INFO:    com.hof.email.starttime=20130911120208
2013-09-11 12:02:08  INFO:    mail.smtp.auth=true
2013-09-11 12:02:08  INFO:    mail.smtp.connectiontimeout=60000
2013-09-11 12:02:08  INFO:    mail.smtp.host=mailserver
2013-09-11 12:02:08  INFO:    mail.smtp.port=25
2013-09-11 12:02:08  INFO:    mail.smtp.timeout=60000
2013-09-11 12:02:08  INFO:    mail.transport.protocol=smtp
2013-09-11 12:02:08  INFO: From: Support
2013-09-11 12:02:08  INFO: To: Customer
2013-09-11 12:02:08  INFO: Subject: Error
2013-09-11 12:02:08  INFO: Message ID: <id>
2013-09-11 12:02:09  INFO: Email sent successfully
2013-09-11 12:02:09  INFO: Javamail session ended
2013-09-11 12:02:09  INFO: ------------------------------

awk '/---/ {if (p) {for (j=0;j<i;j++) print a[j]};i=0;p=0;delete a;a[i++]=$0} !/---/ {a[i++]=$0} /Your/ {p=1}'
2013-09-11 12:02:08  INFO: ------------------------------
2013-09-11 12:02:08  INFO: Javamail session sending email
2013-09-11 12:02:08  INFO: Session properties:
2013-09-11 12:02:08  INFO:    com.hof.email.starttime=20130911120208
2013-09-11 12:02:08  INFO:    mail.smtp.auth=true
2013-09-11 12:02:08  INFO:    mail.smtp.connectiontimeout=60000
2013-09-11 12:02:08  INFO:    mail.smtp.host=mailserver
2013-09-11 12:02:08  INFO:    mail.smtp.port=25
2013-09-11 12:02:08  INFO:    mail.smtp.timeout=60000
2013-09-11 12:02:08  INFO:    mail.transport.protocol=smtp
2013-09-11 12:02:08  INFO: From: Support
2013-09-11 12:02:08  INFO: To: Customer
2013-09-11 12:02:08  INFO: Subject: Your Report Data
2013-09-11 12:02:08  INFO: Message ID: <id>
2013-09-11 12:02:09  INFO: Email sent successfully
2013-09-11 12:02:09  INFO: Javamail session ended

Answer 3

为了好玩，这里有一个GNU grep方法，通过多行搜索来解决这个问题。有关此方法的详细信息great answer

grep -ozP '(?s)(?<=--\n).*?Subject: Your Report Data.*?(?=\n\N*?--)'

Answer 4

对于不同数量的行，您也可以使用此Ruby代码：

ruby -e 'exp = Regexp.new("^[^\n]+INFO: -{30}$.*?INFO: Subject: #{Regexp.escape(ARGV.shift)}$.*?-{30}$", Regexp::MULTILINE); File.read(ARGV.shift).scan(exp).each{|e| puts e}' "Your Report Data" file

它不会将主题中的字符解释为正则表达式字符。

在任何行匹配第二个模式的模式之间打印所有行

4 个答案: