how to get line number of different words if exists in same line

Shell Programming and Scripting
1

I have some txt files. I have to create another text file which contains the portion starting from the format "Date Sex Address" to the end of the file. While using grep -n on Date it also gives me the previous line containg Date. and also Date may be DATE in some files.
My file is like this
Name:
Mod Date
Mod By
Description
Date Sex Address
...
....
...

I want my new file to contain from "Date sex " to the end of file. thanks in advance.

2

Something like this should do it:

awk '/^Date Sex Address/{f=1;next}f' file > newfile

Regards

3

Hi,
I need the line number which contains these three words if contain in the same line.

4

Like this?

awk '/^Date Sex Address/ || /^DATE Sex Address/ {print NR}' infile
5

Assuming that the pattern is "Date Sex Address" and not any combination thereof

$ sed -n '/Date Sex Address/=' file
5
6

I want the line number from the file which contains all the 3 words using any unix command or utility. i am intrested in getting the line number

7 
grep -n 'Date Sex Address' file

Regards

8

Thanks Franklin,
but the space between the words is not consistent. Is there any method to overcome the no of spaces. Some where the Date is written as DATE and I need those also

9 
grep -ni "Date.*Sex.*Address" file
10

I have used egrep -n "Date|Sex|Address" and it is not working where Date is occuring multiple times. Please suggest what to use egrep pattern1 and pattern2

11

Post your input file in the original format and the exact output you want.
Please place them within code tags (select the text and click on the # symbol above the edit window).