在Linux中使用AWK将日期转换为纪元时间 [英] Convert date to epoch time using AWK in linux

问题描述

我有一个逗号分隔的文件，第一列的日期为01/31/2010 我想更改为时代，例如文件"file.csv":

I have a comma-separated file, with the first column as a date of the format 01/31/2010 that I want to change into epoch time, such that the file "file.csv":

 01/30/2010,1,"hi"
 01/31/2010,3,"bye"

将更改为"output.csv":

will change into "output.csv":

 1264809600,1,"hi"
 1264896000,3,"bye"

我知道命令行日期-d"01/30/2010" +％s可以使用，但是只能在一个日期上使用，我需要将其输入到表中，因此，有没有办法使用awk带有一些func():

I know the command line date -d "01/30/2010" +%s will work, but only on a single date, and I need to feed it into a table, so, is there a way to use awk with some func():

cat file.csv | awk -F, 'print func($1)","$2","$3}'

由于我不太在意如何执行此操作，因此，当字符串为mm/dd/yyyy ...时，如何将Excel中的日期更改为纪元...

Since I don't really care how I do this, alternatively, how would I change a date in excel into epoch, when the string is mm/dd/yyyy...

推荐答案

TZ=PST awk -F, '{split($1,date,"/");
                 $1=mktime(date[3] " " date[1] " " date[2] " " "00 00 00");
                 print}'

或者，调用date:

TZ=PST awk -F, '{ OFS = FS;
                  command="date -d" $1 " +%s";
                  command | getline $1;
                  close(command);
                  print}'

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