Print all lines after a string

Shell Programming and Scripting
1

My file has the following contents...

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...

24-JUL-2014 15:04

Thanks

2

Please use

```text
 and 
```

tags around any data or programs.

Try this:

sed 's/^.*LogonTime://' infile
3

The query gives me the following output...

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

User:SYS,O/SUser:oracle,Process:10813484,Machine:host123.xxx.xxx.xxxxxxx.com.au
24-JUL-2014
15:04
4 
sed 's/.*LogonTime:\(.*\)/\1/' file

awk -F 'LogonTime:' 'NF>1{print $2}' file
5

Or try:

awk -F"LogonTime:" '{print $2}' infile

[/FONT][/COLOR]

6

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:

awk '
f {	printf("%s", $0)
	next
}
{	if(f = sub(/.*LogonTime:/, ""))
		printf("%s", $0)
}
END {	if(f)
		print ""
}' file
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