Separate date timestamp use awk or sed command ?

Shell Programming and Scripting
1

Hi,

I have logfile like this :

Actually the format is date format :
yyyymmddHHMMSS

and i want the log become this format
yyyy-mm-dd HH:MM:SS

for example
2009-07-19 11:46:52

Can somebody help me ?
Thanks in advance

2

try..

awk '{print substr($1,1,4)"-"substr($1,5,2)"-"substr($1,7,2),substr($1,9,2)":"substr($1,11,2)":"substr($1,13,2)}' file
3

One way to do it in perl:

$ 
$ perl -ne '@x=unpack("A4 A2 A2 A2 A2 A2",$_); printf("%s-%s-%s %s:%s:%s\n",$x[0],$x[1],$x[2],$x[3],$x[4],$x[5])' logfile.txt
2009-05-13 12:34:15
2009-07-19 14:01:36
2009-07-12 09:23:40
2009-07-25 07:49:55
2009-07-25 07:49:56
2009-07-25 07:49:55
2009-07-25 07:49:57
2009-07-19 10:56:25
2009-07-12 09:23:40
2009-07-12 09:23:40
2009-07-24 19:11:30
2009-07-19 11:09:32
2009-07-12 09:23:40
2009-07-25 13:35:51
2009-07-19 12:26:39
2009-07-12 09:23:40
2009-07-25 14:22:50
2009-07-24 22:31:47
2009-07-19 11:46:52
$

tyler_durden

4

sed line will be shorter:

sed 's/\(....\)\(..\)\(..\)\(..\)\(..\)\(..\)/\1-\2-\3 \4:\5:\6/' logfile.txt