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How to identify delimiter to find and replace a string with sed?


 
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# 15  
RudiC:

Thanks. I debugged as this, this time, sed found the specific MON and replace it with whole value of $NEWDATE. But the DD and YYYY are not searched and replaced. I put DD in search, then the command doesn't work again. Here is output,

Code:
/home/oracle> /usr/xpg4/bin/sed "s/\(JUL\)/$NEWDATE/" test4.txt
       ep.begin_date, ep.end_date, ep.facility_code,
AND    ep.begin_date <= '01-01-OCT-2019-2019'
       ep.begin_date, ep.end_date, ep.facility_code,
AND    ep.begin_date <= '01-01-OCT-2019-2019'
       ep.begin_date, ep.end_date, ep.facility_code,
AND    ep.begin_date <= '01-01-OCT-2019-2019'

/home/oracle> /usr/xpg4/bin/sed "s/[0-3][0-9]\(JUL\)/$NEWDATE/" test4.txt
       ep.begin_date, ep.end_date, ep.facility_code,
AND    ep.begin_date <= '01-JUL-2019'
       ep.begin_date, ep.end_date, ep.facility_code,
AND    ep.begin_date <= '01-JUL-2019'
       ep.begin_date, ep.end_date, ep.facility_code,
AND    ep.begin_date <= '01-JUL-2019'

I will continue debug DD and YYYY
Thanks
# 16  
RudiC:

I have tried different ways. it didn't work. So I will take stupid way to do this work. I will pick up $OLDDATE and $NEWDATE two variables from database. Then:

Code:
/usr/xpg4/bin/sed "s/$OLDDATE/$NEWDATE/" test4.txt > test4_output.txt

It sounds working. Then I go to next step. Thanks anyway for your help and advice. I will continue to learn from this forum.

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