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awk to print all lines after a pattern is found

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# 1  
Old 11-20-2012
awk to print all lines after a pattern is found
Is there a way with aw to print all lines after a string is found

There is a file like this
Code:
.......
........
2012/19/11 :11.58 PM some data
lne no date
2012/19/11 :11.59 PM some other data
2012/20/11 :12.00 AM some other data
some line without dates
some more lines without dates
2012/20/11 :12.01 AM some more data
....
.......

I need a way to display all lines after the first occurrence of the string 2012/20/11

That is

Code:
2012/20/11 :12.00 AM some other data
some line without dates
some more lines without dates
2012/20/11 :12.01 AM some more data
....
.......

# 2  
Old 11-20-2012
print section of file from regular expression to end of file
Code:
awk '/2012\/20\/11/,EOF { print $0 }' filename

Last edited by Franklin52; 11-20-2012 at 05:42 AM.. Reason: Please use code tags for data and code samples
This User Gave Thanks to subramanian For This Post:
swayam123
# 3  
Old 11-20-2012
Hi

Code:
awk '/2012\/20\/11/{f=1;}f' file

Guru.
# 4  
Old 11-20-2012
Code:
sed -n '\|2012/20/11|,$p' infile

# 5  
Old 11-20-2012
Substitute your pattern for PAT:
Code:
awk '/PAT/,0' file

Regards,
Alister
# 6  
Old 11-20-2012
How can I pass the date string pattern as variable
I am trying to run this in a shell script.
In the following way

Code:
awk -v x="11\/20\/12" '/x/,EOF{print $0}' datefiltertestfile.txt

This is not working
# 7  
Old 11-20-2012
Code:
awk -v x='11/20/12' '$0~x,EOF' datefiltertestfile.txt

This User Gave Thanks to elixir_sinari For This Post:
swayam123
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