How to output a shell command using echo

Question

I am trying to format a shell out

str="set @face_date = '20130612'

The below code returns the out as '20130612'

echo $str |  cut -d '=' -f2

But the following code is not working

echo "face_date:=TO_DATE("$str | cut -d '=' -f2"),'YYYY/MM/DD');"

Expected Result:

face_date:= to_date('2013-06-12','YYYY/MM/DD');

Thanks in advance for your help

You're nesting double-quotes inside other double-quotes. Try escaping them with backslashes. — Tom Zych
– Tom Zych, Commented Jun 25, 2014 at 21:40
perhaps try an embedded echo in backquote. not sure how newlines are handled though. except for that should work. like echo "face_date:=TO_DATE("`echo $str | cut -d '=' -f2`"),'YYYY/MM/DD');" disclaimer: haven't tried and i seldom use Unix-land shells. — Cheers and hth. - Alf
– Cheers and hth. - Alf, Commented Jun 25, 2014 at 21:40

anubhava · Accepted Answer · 2014-06-25 21:55:50Z

1

You can use:

echo "face_date:=TO_DATE($(cut -d '=' -f2 <<< "$str")),'YYYY/MM/DD');"

OUTPUT:

face_date:=TO_DATE( '20130612'),'YYYY/MM/DD');

answered Jun 25, 2014 at 21:55

anubhava

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Comments

Sandeep · Accepted Answer · 2014-06-26 11:00:50Z

0

One more way,

echo "face_date:=TO_DATE($(echo $str | cut -d '=' -f2)),'YYYY/MM/DD');"

This will produce following output.

face_date:=TO_DATE( '20130612'),'YYYY/MM/DD');

or you can use

str1=$(echo $str |  cut -d '=' -f2)
echo "face_date:=TO_DATE($(echo ${str1:0:6}-${str1:6:2}-${str1:8:3}),'YYYY/MM/DD');"

for following output

face_date:=TO_DATE('2013-06-12','YYYY/MM/DD');

answered Jun 26, 2014 at 11:00

Sandeep

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Comments

Community · Accepted Answer · 2017-03-20 10:18:25Z

0

Maybe that can help you. You need to include your command in ` (ASCII #96) in order to get executed. Perhaps that question is a better fit here though.

edited Mar 20, 2017 at 10:18

CommunityBot

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answered Jun 25, 2014 at 21:44

Michael

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Comments

David C. Rankin · Accepted Answer · 2014-06-25 22:01:29Z

0

You need to either escape the inner double quotes, or surround in $() or ``:

echo "face_date:=TO_DATE($("$str | cut -d '=' -f2")),'YYYY/MM/DD');"

or simply get rid of all the pipes and cut, just do:

echo "face_date:=TO_DATE(${str##* }),'YYYY/MM/DD');"

edited Jun 25, 2014 at 22:01

answered Jun 25, 2014 at 21:54

David C. Rankin

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