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Grep pattern and display all lines below


 
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# 1  
Old 06-17-2014
Grep pattern and display all lines below

Hi I need to grep for a patter and display all lines below the pattern.

For ex: say my file has the below lines

file1
file2
file3
file4
file5

I NEED to grep for patter file3 and display all lines below the pattern. do we have an option to get this data. Let me know if you require any additional information on the same.
# 2  
Old 06-17-2014
try
Code:
nawk '{A[++c] = $0} END { for ( i = 1; i <=c; i++ ) {if(A[i] ~ "file3") {d++} { if ( d  == 1) print A[i+1]}}}' filename

Also
Code:
awk '/file3/{p++;next} p==1' filename


Last edited by Makarand Dodmis; 06-17-2014 at 06:27 AM..
# 3  
Old 06-17-2014
Try
Code:
sed -n '/file3/,$p' file

Code:
awk '/file3/{p=1}p==1' file

?
# 4  
Old 06-17-2014
He wants all line below pattern "file3" by your code it is displaying "file3" as well
This User Gave Thanks to Makarand Dodmis For This Post:
# 5  
Old 06-17-2014
Try
Code:
awk 'p; /file3/{p=1}' file4

# 6  
Old 06-17-2014
Code:
$ awk 'f; !f{f=/file3/}' <<EOF
file1
file2
file3
file4
file5
EOF

file4
file5


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