Help with Passing the Output of grep to sed command - to find and replace a string in a file. Post: 302913710

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Top Forums Shell Programming and Scripting Help with Passing the Output of grep to sed command - to find and replace a string in a file. Post 302913710 by SriRamKrish on Tuesday 19th of August 2014 12:45:11 PM

08-19-2014

Registered User

Quote:

Originally Posted by balajesuri

Code:

[user@host ~]$ x=20140819; sed "s/=[0-9]\{8\}/=$x/" file
SomeTextGoesHere
[ASD.EXAMPLE@ABCD]
$$TODAY_DT=20140819

[FGH.EXAMPLE@ABCD]
$$TODAY_DT=20140819

[QWE.EXAMPLE@ABCD]
$$TODAY_DT=20140819

Quote:

Originally Posted by Yoda

Code:

sed "s#\(.*=\)\(.*\)#\1$(date +%Y%m%d)#" file

Quote:

Originally Posted by RavinderSingh13

Hello,

Could you please try the following code, hope this helps in variable named s1 you can set the date value which you want to get in outpput.

Code:

awk -F"=" -vs1="20140827" '/^\$\$TODAY/ {$2=s1} 1' OFS="="  Filename

output will be as follows.

Code:

SomeTextGoesHere
[ASD.EXAMPLE@ABCD]
$$TODAY_DT=20140827

[FGH.EXAMPLE@ABCD]
$$TODAY_DT=20140827

[QWE.EXAMPLE@ABCD]
$$TODAY_DT=20140827

If you want to give today's date as passing variable then following may help.

Code:

awk -F"=" -vs1="$(date +%Y%m%d)" '/^\$\$TODAY/ {$2=s1} 1' OFS="="  filename

Thanks,
R. Singh

Solved. You people are my heroes!! Thanks!!

Cheers.
Smilie

SriRamKrish

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