bash 如何使用时间戳进行 grep 计数
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/13955466/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How can i do a grep count by using timestamp
提问by user1916191
How can I do a grepcount by using timestamp?
如何grep使用时间戳进行计数？
Example: If I have a file in which I search a value xyzeverytime. The file gets updated regularly.
示例：如果我有一个文件，我xyz每次都在其中搜索一个值。该文件会定期更新。
20121912-07:15:55 abc cbfr xyz
20121912-07:16:40 mni cbfr xyz
-----------
-----------
-----------
20121912-08:15:55 gty cbfr xyz
20121912-08:20:55 jui uio xyz
I want to find out the occurences of xyzafter 20121912-08:15:55which in this case should be 2.
我想找出的出现次数xyz后，20121912-08:15:55在这种情况下应该是2。
Doing a grep -c "xyz" filenamereads the entire file and gives the result. I want to do it after the last update or using a timestamp.
执行 agrep -c "xyz" filename读取整个文件并给出结果。我想在上次更新或使用时间戳之后执行此操作。
回答by Kent
try this one-liner:
试试这个单线：
awk '$NF=="xyz"&&>="20121912-08:15:55"{x++;}END{print x}' file
回答by Steve
I'm assuming you want to find the occurrences of pattern: 'xyz' where the date/time value is greater than or equal to a specified date/time: '20121912-08:15:55'. Here's what I'd do using GNU awk. Run like:
我假设您想找到模式的出现：'xyz'，其中日期/时间值大于或等于指定的日期/时间：'20121912-08:15:55'。这是我会使用GNU awk. 运行如下：
awk -v pattern="xyz" -v time="20121912-08:15:55" -f script.awk file
Contents of script.awk:
内容script.awk：
BEGIN {
stamp = convert(time)
}
2
~ pattern && convert() >= stamp {
i++
}
END {
print i
}
function convert(var) {
x = "(....)(..)(..)-(..):(..):(..)"
y = "\1 \3 \2 \4 \5 \6"
return mktime(gensub(x,y,"",var))
}
Results:
结果：
awk -v pattern="xyz" -v time="20121912-08:15:55" 'BEGIN { stamp = convert(time) } ts="20121912-08:15:55" patt="xyz" perl -lane '
BEGIN {
($wanted_ts = $ENV{ts}) =~ s/^(....)(..)(..)//;
$pattern = qr{$ENV{patt}};
}
($this_ts = $F[0]) =~ s/^(....)(..)(..)//;
$count++ if $this_ts ge $wanted_ts and /$pattern/;
END {print $count}
'
~ pattern && convert() >= stamp { i++ } END { print i } function convert(var) { return mktime(gensub(/(....)(..)(..)-(..):(..):(..)/,"\1 \3 \2 \4 \5 \6","",var)) }' file
Alternatively, here's the one-liner:
或者，这是单线：
$ sed -n '/20121912-08:15:55/,$p' input.txt | grep -c xyz
回答by glenn Hymanman
Taking inspiration from Kent's answer, here's some Perl that manipulates the odd timestamp into YYYYMMDD format:
从 Kent 的回答中汲取灵感，这里有一些 Perl 将奇数时间戳处理为 YYYYMMDD 格式：
$ fgrep -A 100 '20121912-08:15:55' file | fgrep -c 'xyz'
2
回答by holygeek
You can tell sed to print lines from a file given a range (start and stop point) - the range can be regex or line number notation.
您可以告诉 sed 从给定范围（起点和终点）的文件中打印行 - 范围可以是正则表达式或行号表示法。
For your need this should do it:
根据您的需要，这应该这样做：
$ sed -n '/20121912-08:15:55/,$p' file | fgrep -c 'xyz'
2
Here the start point is given by the date, treated as a regular expression and the end point is the last line symbol $. ptells sed to print the lines within the range given. The -noption to sed tells it to not print the lines that it is processing.
这里的起点由日期给出，被视为正则表达式，终点是最后一行符号$。p告诉 sed 打印给定范围内的行。-nsed的选项告诉它不打印它正在处理的行。
回答by Chris Seymour
This is kind of a hack but just grepfor the earliest date you want and print all lines after that using -Aand then pipe to grep -c xyz:
这是一种黑客行为，但仅grep针对您想要的最早日期-A，然后使用然后通过管道打印所有行grep -c xyz：
$ sort file | sed -n '/20121912-08:15:55/,$p' | fgrep -c 'xyz'
2
Note: fgrepis just fixed string grepas you're not using regexpatterns, it's the same as doing grep -F.
注意：fgrep只是固定字符串，grep因为您没有使用regex模式，它与执行grep -F.
As less hacky way would be to use sedto print all lines from the date, this way you wouldn't need to make sure the value to -Awould cover the length of the file:
由于使用不那么笨拙的方法sed来打印日期中的所有行，这样您就不需要确保 to 的值-A将覆盖文件的长度：
grep xyz filename | sed -r 's/^([^ ]+).*/ 20121912-08:15:55 <= /' | sed -r 's/([0-9]{4})([0-9]{2})([0-9]{2})//g' | sed 's/[-:]//g' | bc | grep 1 | wc -l
This assumes of course you file is in sorted order by timestamps if it's not then:
这当然假设您的文件按时间戳排序，如果不是那么：
grep xyz filename //gets all interseting lines
| sed -r 's/^([^ ]+).*/ 20121912-08:15:55 <= /' //transform them into
//comparison with the
//date you want
| sed -r 's/([0-9]{4})([0-9]{2})([0-9]{2})//g' //invert day and month
| sed 's/[-:]//g' //remove separators
| bc //ask bc result
| grep 1 //get true results only
| wc -l //and finally count them
回答by psycho
Hmmm, quickly written one :
嗯，快写一篇：
20121912-08:20:55 jui uio xyz //grep 'xyz'
20121912-08:15:55 <= 20121912-08:20:55 //sed
20121219-08:15:55 <= 20121219-08:20:55
20121219081555 <= 20121219082055
1 //result from bc
It's pretty ugly (I'm not a sed nor command line master) and may probably be shortened, but it's a way to do it. Explanation below :
它非常丑陋（我不是 sed 也不是命令行大师）并且可能会被缩短，但这是一种方法。解释如下：
##代码##For last line of your example, the steps would give :
对于示例的最后一行，步骤将给出：
##代码##HTH
HTH
