Question

我的数据是：

 - List item

 - COL_DPU_JA08 Cash Remit 3/09/2017 ILA WSE
 - COL_DPU_MBTC_CJPM_03.09.17
 - COL_DPU_CJOA_CA_3.9.17
 - COL_DPU_CJNA_CA_03/09/2017
 - COL_POS_CJDB_BDO_03092017
 - COL_DPU_JE12 Cash Remit 3/09/2017 TUG WSE
 - COL_DPU_JA08 Checks Remit 3/10/17 ILA

如何提取日期并将所述日期转换为以下格式：的 MM.DD.YY

Answer 1

由于您没有提供您正在使用的语言，因此这是一个执行此任务的perl脚本：

#!/usr/bin/perl
use Modern::Perl;
use Data::Dump qw(dump);

my $re = qr~[ _](\d\d?)\D?(\d\d?)\D?(?:\d\d)?(\d\d)\b~;
while(<DATA>) {
    chomp;
    my (@l) = $_ =~  $re;
    dump@l;
}

__DATA__
 - COL_DPU_JA08 Cash Remit 3/09/2017 ILA WSE
 - COL_DPU_MBTC_CJPM_03.09.17
 - COL_DPU_CJOA_CA_3.9.17
 - COL_DPU_CJNA_CA_03/09/2017
 - COL_POS_CJDB_BDO_03092017
 - COL_DPU_JE12 Cash Remit 3/09/2017 TUG WSE
 - COL_DPU_JA08 Checks Remit 3/10/17 ILA

<强>输出：

(3, "09", 2017)
("03", "09", 17)
(3, 9, 17)
("03", "09", 17)
("03", "09", 17)
(3, "09", 17)
(3, 10, 17)

正则表达式：如何提取不同格式和地点的日期

1 个答案: