How to divide digit on a condition and to unite through a divider?

Question

There is a variable which contains 20190105221953.ogg,
Want it turns to "2019-01-05 19:59:53"

Tried to apply regex with needed quantity of numbers grep -Eo '[[:digit:]]{4}'
But It's not working.

Where did the `19:59:50` come from :O? – PesaThe Jan 05 '19 at 15:40 — PesaThe, Jan 05 '19 at 15:40
You can always use substring: `"${var:0:4}-${var:4:2}..."`. – PesaThe Jan 05 '19 at 15:46 — PesaThe, Jan 05 '19 at 15:46

score 1 · Answer 1 · answered Jan 05 '19 at 16:34

1

You can try converting your string to datetime with date command:

date -d "${var:0:8} ${var:8:2}:${var:10:2}:${var:12:2}" +"%Y-%m-%d %H:%M:%S"

This may help for flexible output of the date by changing "%Y-%m-%d %H:%M:%S"

answered Jan 05 '19 at 16:34

Ardit

1,522
17
23

score 0 · Accepted Answer · answered Jan 05 '19 at 16:06

0

Thanks, PesaThe!

var=`date +"%Y%m%d%H%M%S"`.ogg
echo ${var:0:4}-${var:4:2}-${var:6:2} ${var:8:2}:${var:10:2}:${var:12:2}

answered Jan 05 '19 at 16:06

Anton Teremshonok

7
1
2

Consider maybe shorter: `var=$(date +"%F %T"); echo "${var} >> ${var//[-: ]}.ogg"`. – PesaThe Jan 05 '19 at 16:11
`var=`date +"%Y%m%d%H%M%S"`.ogg` - this only tested dynamic filename, filename `20190105221953.ogg` does not change. – Anton Teremshonok Jan 05 '19 at 16:15
I see. Thought you might be creating a filename and printing the date in a readable format. – PesaThe Jan 05 '19 at 16:16

score 0 · Answer 3 · answered Jan 05 '19 at 17:46

0

Though it seems awkward to use sed... :

echo $var
#=> 20190105221953.ogg

echo $var|sed -E 's/([0-9]{4})([0-9]{2})([0-9]{2})([0-9]{2})([0-9]{2})([0-9]{2}).*/\1-\2-\3 \4:\5:\6/'
#=> 2019-01-05 22:19:53

answered Jan 05 '19 at 17:46

Til

5,150
13
26
34

How to divide digit on a condition and to unite through a divider?

3 Answers3