date - 将 csv 文件的 yyyymmdd 转换为 mm/dd/yyyy

Question

请帮忙。我需要一个用于 example.csv 文件的 unix shell 命令，其记录如下:

1231,aaa,bbb,20161001,20161002,hi-ax
1231,aaa,ccc,20161002,20161003,hi-ay 
1231,aaa,ddd,20161001,20161007,hi-az

输出应该是:

1231,aaa,bbb,10/01/2016,10/02/2016,hi-ax
1231,aaa,ccc,10/02/2016,10/03/2016,hi-ay
1231,aaa,ddd,10/01/2016,10/07/2016,hi-az

有人可以修复这个命令吗？我已经尝试过，但如果没有分隔符，它将无法工作:

awk -F, '{split($4,a,"");$4=a[2]"/"a[3]"/"a[1]}1' OFS=, sample.csv > output.csv

Answer 1

我宁愿使用这样的函数，它使其更易于维护，例如:

convert-date.awk

function ruin_utc(d) {
  year  = substr(d, 1, 4)
  month = substr(d, 5, 2)
  day   = substr(d, 7, 2)
  return month "/" day "/" year
}

您现在可以像这样使用该函数:

awk -f convert-date.awk -e '{ $4 = ruin_utc($4); $5 = ruin_utc($5) } 1' FS=, OFS=, sample.csv

或更便携:

convert-date2.awk

function ruin_utc(d) {
  year  = substr(d, 1, 4)
  month = substr(d, 5, 2)
  day   = substr(d, 7, 2)
  return month "/" day "/" year
}
{ $4 = ruin_utc($4); $5 = ruin_utc($5) } 
1

像这样运行:

awk -f convert-date.awk FS=, OFS=, sample.csv

输出:

1231,aaa,bbb,10/01/2016,10/02/2016,hi-ax
1231,aaa,ccc,10/02/2016,10/03/2016,hi-ay
1231,aaa,ddd,10/01/2016,10/07/2016,hi-az