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print remaining part from the first-match within a file

 
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# 1  
Old 12-23-2008
print remaining part from the first-match within a file
Hi, i was looking for unix command(s) for :

find the first occurrence of a given pattern with in a file and print the remaining part.


below is an example of what i am looking for:

lets say, a file named myfile.txt
Quote:
# more myfile.txt

A
B
C
D
E

#
now, the command i am looking for will do the following

Quote:
# my_command C myfile.txt

C
D
E

#
# 2  
Old 12-23-2008
Hi, If I knew that the number of lines in the source file was limited, I would do it the lazy way,
something like
grep -A9999 C myfile.txt
(meaning: include the 9999 lines following "C" in the output)
but I'm sure someone here soon will supply a solution in sed or awk, it's a common enough question.

/Lakris
# 3  
Old 12-23-2008
Bug
Code:
 sed -n '/C/,${p;}' infile
C
D
E

HTH
# 4  
Old 12-25-2008
Quote:
Originally Posted by Lakris
Hi, If I knew that the number of lines in the source file was limited, I would do it the lazy way,
something like
grep -A9999 C myfile.txt
(meaning: include the 9999 lines following "C" in the output)
but I'm sure someone here soon will supply a solution in sed or awk, it's a common enough question.

/Lakris
thanks.
however, in my case, the grep shows -A as an illegel option.
i m on solaris:

grep: illegal option -- A
Usage: grep -hblcnsviw pattern file . . .
# 5  
Old 12-25-2008
Quote:
Originally Posted by Tytalus
Code:
 sed -n '/C/,${p;}' infile
C
D
E

HTH
thanks a lot ... working fine.
 
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