CODEWITHSUNDEEP
bashshellcsvawksed
Asfbar
Asfbar

Reputation: 279

Sed search and replace in every line in file, first raw

i'm trying to develop a shell script which scan a csv file (example below) and replace the EPOCH time stamp with the a readable time stamp.

the csv file looks like (column is EPOCH) :

1464007935578, 852111
1464007935600, 852111
1464007935603, 852111
1464007935603, 1900681

my script code:

#!/bin/bash

file=xxx.csv
while read line; do
    epoch=`echo $line | awk -F',' '{print $1}'`
    miliSec=$(($epoch % 1000 ))
    timeWithoutMiliSec=`echo "$(date +"%T" -d@$(($epoch / 1000)))"`
    fullTime=`echo "$timeWithoutMiliSec,$miliSec"`
    echo $line | awk -F',' '{print $1}'|sed -i 's/[0-9]*/'$fullTime'/g' $file
done <$file

Desired output

12:52:15,255 852111
12:52:15,257 852111
12:52:15,259 852111
12:52:15,261 1900681

but when i run script it stuck and create many files start with sed* and i must kill the script in order to stop it. i think that the problem might be with the while loop combining with the sed while both (sed and while) take argument $file

can someone please advise about that issue?

Thanks a lot!

Upvotes: 0

Views: 167

Answers (1)

fedorqui
fedorqui

Reputation: 289725

Use awk's strftime() function:

$ awk -F, -v OFS=, '{print strftime("%F", $1/1000), $2}' file
2016-05-23, 852111
2016-05-23, 852111
2016-05-23, 852111
2016-05-23, 1900681

This uses

%F

Equivalent to specifying ‘%Y-%m-%d’. This is the ISO 8601 date format.

Note I have to divide by 1000 because you have the miliseconds in your input.

Upvotes: 2

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