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


 
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# 1  
Help with Passing the Output of grep to sed command - to find and replace a string in a file.

I have a file example.txt as follows :
Code:
SomeTextGoesHere
[ASD.EXAMPLE@ABCD]
$$TODAY_DT=20140818

[FGH.EXAMPLE@ABCD]
$$TODAY_DT=20140818

[QWE.EXAMPLE@ABCD]
$$TODAY_DT=20140818

I need to automatically update the date (20140818) in the above file, by getting the new date as argument, using a shell script.
(It would even be better if I could pass both the file name, in which the date should be updated, and date as arguments)

I tried greping the line where the date occurs using
Code:
grep '[0-9]' a.txt

which gives the output as :
Code:
$$TODAY_DT=20140818
$$TODAY_DT=20140818
$$TODAY_DT=20140818

Now I don't know how to give this output as input to a sed command and replace the date(only date, the numeric part).

Or, is there any other alternative way to achieve this ??

Cheers.

Last edited by rbatte1; 08-19-2014 at 01:50 PM..
# 2  
try
Code:
sed 's/20140818/20140819/g' a.txt > b.txt && mv b.txt a.txt

# 3  
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

This User Gave Thanks to balajesuri For This Post:
# 4  
Code:
sed "s#\(.*=\)\(.*\)#\1$(date +%Y%m%d)#" file

# 5  
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 output.

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

Last edited by rbatte1; 08-19-2014 at 01:52 PM.. Reason: RavinderSingh13 Added one more solution - RBATTE1 for spelling
This User Gave Thanks to RavinderSingh13 For This Post:
# 6  
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

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