Cutting part of line/string in shell scripting

Question

I have the following line:

Jan 13, 2014 1:01:31 AM

I want to remove the seconds part of the line. The result should be:

Jan 13, 2014 1:01 AM

How can this be done ?

It would be better if you indicate how you are obtaining such line, because we could try to change the date output format. — fedorqui
– fedorqui, Commented Jan 14, 2014 at 14:27

choroba · Accepted Answer · 2014-01-14 14:29:49Z

2

Use parameter expansion:

t='Jan 13, 2014 1:01:31 AM'
ampm=${t: -2}               # last two characters
echo "${t%:*} $ampm"        # remove everything after the last :

answered Jan 14, 2014 at 14:29

choroba

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1 Comment

Aleks-Daniel Jakimenko-A. Over a year ago

+1, this is the most efficient way unlike spawning other processes like sed.

anubhava · Accepted Answer · 2014-01-14 14:28:25Z

1

Using sed:

s='Jan 13, 2014 1:01:31 AM'
sed 's/:[0-9]*\( [AP]M\)/\1/' <<< "$s"
Jan 13, 2014 1:01 AM

answered Jan 14, 2014 at 14:28

anubhava

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Comments

Kent · Accepted Answer · 2014-01-14 14:31:04Z

1

you can give this a try:

sed 's/:[^:]* / /'

with your example:

kent$ (master|✚2) echo "Jan 13, 2014 1:01:31 AM"|sed 's/:[^:]* / /'
Jan 13, 2014 1:01 AM

answered Jan 14, 2014 at 14:31

Kent

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BMW · Accepted Answer · 2014-01-16 12:04:44Z

0

Another way, if your date command is gnu date which support -d option.

$ str="Jan 13, 2014 1:01:31 AM"

$ date -d "$str" +"%b %d, %Y %l:%M %p"

Jan 13, 2014  1:01 AM

edited Jan 16, 2014 at 12:04

answered Jan 14, 2014 at 23:47

BMW

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