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Print all lines after a string

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# 1  
Old 07-28-2014
Print all lines after a string
My file has the following contents...

Code:
User:SYS,O/SUser:oracle,Process:10813484,Machine:host123.xxx.xxx.xxxxxxx.com.au,Program:sqlplus@host123.xxx.xxx.xxxxxxx.com.au(TNSV1-V,LogonTime:24-JUL-2014 
15:04

I would like to print all character appearing after the string LogonTime:

My output shoud be...
Code:
24-JUL-2014 15:04

Thanks
Last edited by Don Cragun; 07-28-2014 at 10:07 PM.. Reason: Add CODE tags.
# 2  
Old 07-28-2014
Please use [code] and [/code] tags around any data or programs.

Try this:

Code:
sed 's/^.*LogonTime://' infile

# 3  
Old 07-28-2014
The query gives me the following output...

Code:
sed 's/^.*LogonTime:*//' sample.log

Code:
User:SYS,O/SUser:oracle,Process:10813484,Machine:host123.xxx.xxx.xxxxxxx.com.au
24-JUL-2014
15:04

Last edited by Nagesh_1985; 07-29-2014 at 01:24 AM..
# 4  
Old 07-29-2014
Code:
sed 's/.*LogonTime:\(.*\)/\1/' file

Code:
awk -F 'LogonTime:' 'NF>1{print $2}' file

# 5  
Old 07-29-2014
Or try:
Code:
awk -F"LogonTime:" '{print $2}' infile


# 6  
Old 07-29-2014
If you would have used CODE tags as required by forum rules, people who looked at your post (before it was edited by an administrator), might have realized that your sample input was two lines instead of one, and might have known that the 1st line in your sample input ended with a space character just before the newline line terminator.

This seems to do what you have requested:
Code:
awk '
f {	printf("%s", $0)
	next
}
{	if(f = sub(/.*LogonTime:/, ""))
		printf("%s", $0)
}
END {	if(f)
		print ""
}' file

This User Gave Thanks to Don Cragun For This Post:
Nagesh_1985
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