Unix模式日期時間匹配

Question

我想編輯這一行：

1987,4,12,31,4,1987-12-31 00:00:00.0000000,UA,19977,UA,,631,12197,1219701,31703,HPN,White Plains, NY,NY,36,New York,22,13930,1393001,30977,ORD,Chicago\, IL,IL,17,Illinois,41,756,802,483.2,6,6,0,0,0700-0759,,,,,914,938,600.8,24,24,1,1,0900-0959,0,,0,138,156,,1,738,3,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,US1NJBG0005,US1ILCK0027,,,,,,,,,,,,,1987-12-31 08:09:12.0000000,519494350

我希望輸出是：

1987,4,12,31,4，1987-12-31 00：00：00.000 ，UA，19977，UA ,, 631,12197,1219701,31703，HPN，White Plains，NY，NY，36，New York， 22,13930,1393001,30977，ORD，Chicago \\，IL，IL，17，伊利諾伊州，41,756,802,483.2,6,6,0,0,0700-0759 ,,,,, 914,938,600.8,24,24,1,1,0900 -0959,0，，0,138,156，，1,738,3 ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, US1NJBG0005 US1ILCK0027 1987-12-31 08：09： 12.000，519494350

我想找到以下每種模式： ****-**-** **:**:**.0000000

並刪除最后4位數字（0000），這樣我得到****-**-** **:**:**.000.

如果有幫助，則此日期格式在第6列和n-1列中。

Answer 1

要獲取第六列的值並刪除最后四位數字，可以使用：

awk -F, '{print substr($6, 0, length($6)-4) }'

同樣，可以通過以下方式訪問N-1列：

awk -F, '{print substr( $(NF-1), 0, length($(NF-1))-4) }'

編輯：

要僅替換列中的值，但仍打印所有內容，請使用：

awk 'BEGIN{ FS=","; OFS=","} 
{ $6=substr($6, 0, length($6)-4); 
  $(NF-1)=substr( $(NF-1), 0,length($(NF-1))-4); 
  print $0}'

Answer 2

基於awk的解決方案

格式精美，可移植的腳本：

#!/usr/bin/awk -f
BEGIN {
    FS = ","  # input:  fields are separated by ,
    OFS = "," # output: fields are separated by ,
}

{
    sub(/[0-9][0-9][0-9][0-9]$/, "", $6)      # remove last 4 digits from the 6th column
    sub(/[0-9][0-9][0-9][0-9]$/, "", $(NF-1)) # remove last 4 digits from the n-1 column
    print
}

使用gawk的單行，便攜式性較低的版本：

gawk --re-interval -F , -v OFS=, '{sub("[0-9]{4}$", "", $6); sub("[0-9]{4}$", "", $(NF-1)); print}'

注意 ：傳統awk的正則表達式引擎不支持{n}重復運算符，因此gawk版本3或更早版本需要使用--re-interval運行。 對於其他awk風格，例如nawk ，您需要像從上面的可移植較長腳本中那樣顯式重復正則表達式。

基於sed的解決方案

sed -r 's/^(([^,]*,){5})([^,]+)[0-9]{4},(([^,]*,)*)([^,]+)[0-9]{4}(,[^,]*)$/\1\3\4\6\7/'

（使用GNU sed-4.2.2-6測試）

Answer 3

這是Perl中的解決方案。

更新-編輯以輸出完整的CSV行，其中的時間戳已被截斷的CSV替代

更新2-更新兩個時間戳列，而不僅僅是第一列

#!/usr/bin/env perl

use strict;
use warnings;
use feature 'say';

use Text::CSV;

my $CSV = Text::CSV->new();

while (my $line = readline(STDIN)) {
    $CSV->parse($line) or die "Unable to parse line '$line'";

    my @fields = $CSV->fields();

    for my $f (@fields) {
        $f =~ s/
            ^               # start of string
            (               # start capture to $1
                \d{4} -     # year
                \d{2} -     # month
                \d{2} \s+   # day
                \d{2} :     # hour
                \d{2} :     # minute
                \d{2} [.]   # second
                \d{3}       # milisecond
            )               # end capture to $1
            \d{4}           # unwanted sub-second precision
            $               # end of string
        /$1/gmsx;
    }

    $CSV->combine(@fields);
    say $CSV->string();
}

例如：

alex@yuzu:~$ cat input.txt 
1987,4,12,31,4,1987-12-31 00:00:00.0000000,UA,19977,UA,,631,12197,1219701,31703,HPN,White Plains, NY,NY,36,New York,22,13930,1393001,30977,ORD,Chicago\, IL,IL,17,Illinois,41,756,802,483.2,6,6,0,0,0700-0759,,,,,914,938,600.8,24,24,1,1,0900-0959,0,,0,138,156,,1,738,3,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,US1NJBG0005,US1ILCK0027,,,,,,,,,,,,,1987-12-31 08:09:12.0000000,519494350

alex@yuzu:~$ ./csv.pl < input.txt
1987,4,12,31,4,"1987-12-31 00:00:00.000",UA,19977,UA,,631,12197,1219701,31703,HPN,"White Plains"," NY",NY,36,"New York",22,13930,1393001,30977,ORD,Chicago\," IL",IL,17,Illinois,41,756,802,483.2,6,6,0,0,0700-0759,,,,,914,938,600.8,24,24,1,1,0900-0959,0,,0,138,156,,1,738,3,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,US1NJBG0005,US1ILCK0027,,,,,,,,,,,,,"1987-12-31 08:09:12.000",519494350

在像Debian這樣的Debian系統上，您應該已經擁有Perl，並且可以使用以下命令安裝Text :: CSV：

$ sudo apt-get install libtext-csv-perl

Answer 4

您也可以嘗試使用此GNU sed命令，

$ sed -r 's/^.*,([^,]*)....,.*$/\1/g' file
1987-12-31 08:09:12.000

如果您只想更換，請嘗試一下，

$ sed -r 's/^(.*,)([^,]*)....(,.*)$/\1\2\3/g' file
1987,4,12,31,4,1987-12-31 00:00:00.0000000,UA,19977,UA,,631,12197,1219701,31703,HPN,White Plains, NY,NY,36,New York,22,13930,1393001,30977,ORD,Chicago\, IL,IL,17,Illinois,41,756,802,483.2,6,6,0,0,0700-0759,,,,,914,938,600.8,24,24,1,1,0900-0959,0,,0,138,156,,1,738,3,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,US1NJBG0005,US1ILCK0027,,,,,,,,,,,,,1987-12-31 08:09:12.000,519494350

我想您希望輸出是這樣的，

$ grep -oP '[0-9]{4}-[0-9]{2}-[0-9]{2} [0-9]{2}:[0-9]{2}:[0-9]{2}\....' file
1987-12-31 00:00:00.000
1987-12-31 08:09:12.000

更新：

$ echo '1987,4,12,31,4,1987-12-31 00:00:00.0000000,UA,19977,UA,,631,12197,1219701,31703,HPN,White Plains, NY,NY,36,New York,22,13930,1393001,30977,ORD,Chicago\, IL,IL,17,Illinois,41,756,802,483.2,6,6,0,0,0700-0759,,,,,914,938,600.8,24,24,1,1,0900-0959,0,,0,138,156,,1,738,3,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,US1NJBG0005,US1ILCK0027,,,,,,,,,,,,,1987-12-31 08:09:12.0000000,519494350' | sed -r 's/([0-9]{4}-[0-9]{2}-[0-9]{2} [0-9]{2}:[0-9]{2}:[0-9]{2}\....)..../\1/g'
1987,4,12,31,4,1987-12-31 00:00:00.000,UA,19977,UA,,631,12197,1219701,31703,HPN,White Plains, NY,NY,36,New York,22,13930,1393001,30977,ORD,Chicago\, IL,IL,17,Illinois,41,756,802,483.2,6,6,0,0,0700-0759,,,,,914,938,600.8,24,24,1,1,0900-0959,0,,0,138,156,,1,738,3,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,US1NJBG0005,US1ILCK0027,,,,,,,,,,,,,1987-12-31 08:09:12.000,519494350