Question

我有一个CSV文件，其日期时间格式如下：

8/2/2012 1:34:01 PM

我需要查找此类日期的所有事件（可能是正则表达式）并将其（在文件中）转换为mysql默认日期时间格式（2012-08-02 13:34:01）。

我找到了

date --date="8/2/2012 1:34:01 PM" +"%F %T"

工作正常，我得到了正确的结果。解决方案是找到所有错误格式化的日期，并通过上述日期函数传递它。

CSV文件中的单行如下所示：

"Data 1","Data 2","Data 3","Data 4","100000000100","100000002263","10/31/2012 10:02:03 AM","1/29/1999 1:54:50 PM","11/6/2012 8:47:01 PM","S"

非常感谢！

Answer 1

这对你好吗？

awk -F, -vOFS=','  '{for (i=7;i<=9;i++) {"date --date="$i" +\"%F %T\"" |getline $i;$i="\""$i"\""}}1' file

用那条线测试：

kent$  echo '"Data 1","Data 2","Data 3","Data 4","100000000100","100000002263","10/31/2012 10:02:03 AM","1/29/1999 1:54:50 PM","11/6/2012 8:47:01 PM","S"'|awk -F, -vOFS=','  '{for (i=7;i<=9;i++) {"date --date="$i" +\"%F %T\"" |getline $i;$i="\""$i"\""}}1'
"Data 1","Data 2","Data 3","Data 4","100000000100","100000002263","2012-10-31 10:02:03","1999-01-29 13:54:50","2012-11-06 20:47:01","S"

Answer 2

如果使用perl是一个选项，这应符合您的要求

#! /usr/bin/perl

use strict;
use warnings;

while (my($m, $d, $y, $H, $M, $S, $a) = m!\"(\d+)/(\d+)/(\d+)\s+(\d+):(\d+):(\d+)\s+([ap]m)\"!i) {
    $H += 12 if ($a =~ m/pm/i);
    my $dt = sprintf("%04d-%02d-%02d %02d:%02d:%02d", $y, $m, $d, $H, $M, $S);
    s!!$dt!;
}

用

调用它

perl -i.bak -p convert.pl csv.txt

这会将日期转换为MySQL格式。它还会创建备份文件csv.txt.bak。

Answer 3

我使用带有适当CSV解析器的工具。这是一个红宝石的例子

ruby -e '
  require "csv"
  require "date"

  def format_date(date)
    DateTime.strptime(date, "%m/%d/%Y %r").strftime "%Y-%m-%d %T"
  end

  CSV.foreach ARGV[0] do |row|
    [6,7,8].each {|i| row[i] = format_date row[i]}
    puts CSV.generate_line row, {:force_quotes => true}
  end
' file.csv

根据您的样本输入，此输出

"Data 1","Data 2","Data 3","Data 4","100000000100","100000002263","2012-10-31 10:02:03","1999-01-29 13:54:50","2012-11-06 20:47:01","S"

使用bash将日期时间（12h）转换为CSV文件中的mysql datetime格式

3 个答案: