Question

我想编辑这一行：

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

我希望输出为：

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，纽约，22,13930,1393001,30977，ORD，Chicago \，IL，IL，17，Illinois，41,756,802,483.2,6,6,0,0,00000-0759 ,,,,, 914,938,600.8,24,24,1,1,0900-0959,0，0138156，1,738,3 ,,,,,,,,,,,,,,,,,,,,,,,,,,, ,,,,,,,,,,,,,,,,,,,,,,,,,,, US1NJBG0005，US1ILCK0027 ,,,,,,,,,,,,,的 1987-12- 31 08：09：12.000 ，519494350

我想找到每个模式：****-**-** **:**:**.0000000

并删除最后4位数字（0000），以便我得到****-**-** **:**:**.000.

如果有用，则此日期格式位于第6列和n-1列。

Answer 1

要获取第6列的值并删除可以使用的最后四位数字：

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

同样，可以通过以下方式访问N-1列：

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

修改

要仅替换列中的值，但仍打印所有内容：

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

基于Awk的解决方案

格式良好的便携式脚本：

#!/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的单行，不太便携的版本：

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

NB 传统 awk 的正则表达式引擎不支持{n}重复运算符，因此 gawk 版本3需要使用--re-interval运行或更早版本。对于 awk 的其他风格，例如 nawk ，您需要明确重复正则表达式，就像上面的便携式长脚本一样。

基于sed的解决方案

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

（使用GNU sed-4.2.2-6测试）

Answer 3

这是Perl的解决方案。

更新 - 编辑输出完整的CSV行，时间戳替换为截断的

更新2 - 更新两个时间戳列，而不仅仅是第一个

#!/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

在类似Debian的系统（如Ubuntu）上，您应该已经拥有Perl，并且可以使用以下命令安装Text :: CSV。

$ sudo apt-get install libtext-csv-perl

Answer 4

你也可以尝试这个GNU sed命令，

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

如果您只想更换，请尝试此操作，

$ 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

我认为您希望输出像这样，

$ 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

<强>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

Unix模式日期时间匹配

4 个答案:

基于Awk的解决方案

基于sed的解决方案