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Convert UNIX timestamp to readable format in the file

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# 1  
Old 09-15-2015
Convert UNIX timestamp to readable format in the file
Hello I have a file : file1.txt with the below contents :

Code:
237176 test1 test2 1442149024
237138 test3 test4 1442121300
237171 test5 test7 1442112823
237145 test9 test10 1442109600

In the above file fourth field represents the timestamp in Unix format.

I found a command which converts this into actual timestamp :

Code:
cat file1.txt | awk '{print $4}' | perl -pe 's/([\d]{10})/localtime $1/eg;'

Code:
Sun Sep 13 08:57:04 2015
Sun Sep 13 01:15:00 2015
Sat Sep 12 22:53:43 2015
Sat Sep 12 22:00:00 2015

How can i convert the Unix timestamp to the actual timestamp and replace it with the fourth field in the file. The Output should look like below :

Code:
237176 test1 test2 Sun Sep 13 08:57:04 2015
237138 test3 test4 Sun Sep 13 01:15:00 2015
237171 test5 test7 Sat Sep 12 22:53:43 2015
237145 test9 test10 Sat Sep 12 22:00:00 2015

Please help. Thanks
# 2  
Old 09-15-2015
Hello rahul2662,

Following may help you in same.
Code:
awk '{ printf "%d %s %s  %s\n", $1, $2, $3, strftime("%c",$NF) }' Input_file

Output will be as follows.
Code:
237176 test1 test2  Sun Sep 13 08:57:04 2015
237138 test3 test4  Sun Sep 13 01:15:00 2015
237171 test5 test7  Sat Sep 12 22:53:43 2015
237145 test9 test10  Sat Sep 12 22:00:00 2015

Thanks,
R. Singh
These 2 Users Gave Thanks to RavinderSingh13 For This Post:
rahul2662 rbatte1
# 3  
Old 09-15-2015
How about
Code:
while read A B C TM; do LC_ALL=C printf "%s %s %s %(%c)T\n" $A $B $C $TM; done <file
237176 test1 test2 Sun Sep 13 14:57:04 2015
237138 test3 test4 Sun Sep 13 07:15:00 2015
237171 test5 test7 Sun Sep 13 04:53:43 2015
237145 test9 test10 Sun Sep 13 04:00:00 2015

(needs a recent shell)
This User Gave Thanks to RudiC For This Post:
rahul2662
# 4  
Old 09-17-2015
Thanks Ravinder and RudiC. It worked fine.

---------- Post updated at 07:11 PM ---------- Previous update was at 06:54 PM ----------

One more question :

Suppose I have a file file1.txt with the below contents :

Code:
a b c e
f g h i j
m y t
EFFECTIVE_TIME 1373938000
l r t o

The second field on Line number 4 is time ( this could be on any row but will always have EFFECTIVE_TIME as first column) which I want to convert into human readable time. So how can we convert this into real time.

Thanks for your help.
# 5  
Old 09-17-2015
Hello Rahul,

You could try as follows for same, let me know if this helps you.
Code:
awk '/EFFECTIVE_TIME/ { printf "%s %s\n", $1, strftime("%c",$NF); next} 1'  Input_file

Output will be as follows.
Code:
a b c e
f g h i j
m y t
EFFECTIVE_TIME Mon Jul 15 21:26:40 2013
l r t o

Thanks,
R. Singh
This User Gave Thanks to RavinderSingh13 For This Post:
rahul2662
# 6  
Old 09-17-2015
Hello Ravinder what does highlighted "1" means in the below script ?

Code:
awk '/EFFECTIVE_TIME/ { printf "%s %s\n", $1, strftime("%c",$NF); next} 1'  Input_file

# 7  
Old 09-17-2015
Hello Rahul,

Following is the explanation for same.
Code:
awk '/EFFECTIVE_TIME/                                   #### Searching for string EFFECTIVE_TIME
 { printf "%s %s\n", $1, strftime("%c",$NF);             #### if above condition is TRUE and a line has string EFFECTIVE_TIME then print the first column and chage last column to actual human readbable time.
 next}                                                   #### Now leave all coming statments/actions by calling next here
 1                                                      #### awk works on method of conditions and then perform actions, by 1 we are making condition as TRUE and not mentioning the action here so awk is performing the default action which is print so it will print al lthe lines
'  Input_file                                           #### mentioning Input_file here.

Thanks,
R. Singh
Last edited by RavinderSingh13; 09-17-2015 at 11:50 AM..
This User Gave Thanks to RavinderSingh13 For This Post:
rahul2662
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