CODEWITHSUNDEEP
bashawk
vujke
vujke

Reputation: 493

BASH variable and awk inside variable

I got this simple bash script:

#! /bin/bash

DATE=$(date +"%d%m%Y%H%M")

DATETIME=$DATE | awk '{print substr($1,1,2)"."substr($1,3,2)"."substr($1,5,4)", "substr($1,9,2)":"substr($1,11,2)}'

echo $DATETIME

The output of variable DATE is e.g.

250520141600

and the output of variable DATETIME should be:

25.05.2014, 16:00

but I got nothing.

What's wrong?

Upvotes: 1

Views: 127

Answers (3)

Tom Fenech
Tom Fenech

Reputation: 74596

There's no need to use awk to format the output. Just do this:

$ DATETIME=$(date +"%d.%m.%Y, %H:%M")
$ echo "$DATETIME"
25.05.2014, 15:21

That said, you might want to store one date and display it in a number of different ways. To do that, I would recommend using date, rather than awk:

$ DATE=$(date)
$ echo "$DATE"       
Sun May 25 15:39:09 BST 2014
$ date -d "$DATE" +"%d%m%Y%H%M"
250520141539
$ date -d "$DATE" +"%d.%m.%Y, %H:%M"
25.05.2014, 15:39

as well as displaying the current date/time, date can also be passed a string containing a date and reformat it for you.

Upvotes: 4

timgeb
timgeb

Reputation: 78650

It is not working because you did not echo the content of DATE before piping it to awk. 

>> DATE=$(date +"%d%m%Y%H%M")
>> DATETIME=$(echo $DATE | awk '{print substr($1,1,2)"."substr($1,3,2)"."substr($1,5,4)", "substr($1,9,2)":"substr($1,11,2)}')
>> echo $DATETIME
25.05.2014, 16:07

Upvotes: 2

apmasell
apmasell

Reputation: 7153

You need to invoke a command and put the output in the variable. Since the output has spaces, it should be quoted.

DATETIME="$(echo $DATE | awk ...)"

Also, why not just adjust the date output to be the format you want?

Upvotes: 3

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