Match multiple regular expressions from a single file using awk

Try with:

awk '/foo/||/bar/' Input.txt

awk regexps are extended regexps while grep's without -E are basic regexp. With extended regexp:

awk '/name=|age=|class=|marks=/{nr[NR]; nr[NR+2]}; NR in nr'

Note that standard basic regexp do not have an alternation operator, so

grep 'a\|b'

Will typically not work in every grep (while a few like GNU grep support it as an extension).

grep -E 'a|b'
grep -e a -e b
grep 'a
b'

Will work in every grep though.

Using grep

What if you used the after context switch to grep (-A) and specified a 1 to get the first line after a match?

$ grep -E -A 1 "name=|age=|class=|marks=" student.txt

Example

Sample file.

$ cat student.txt 
name=
1st line after name
2nd line after name
age=
1st line after age
2nd line after age
class=
1st line after class
2nd line after class
marks=
1st line after marks
2nd line after marks

Then when you execute the above command:

$ grep -E -A 1 "name=|age=|class=|marks=" student.txt
name=
1st line after name
--
age=
1st line after age
--
class=
1st line after class
--
marks=
1st line after marks

Using awk

As @RahulPatil suggested using the construct to awk:

'/string1/||/string2/||...'

Something like this would do what you're looking for.

$ awk '
  /name=/||/age=/||/class=/||/marks=/{nr[NR]; nr[NR+1]}; NR in nr
' student.txt 

Example

$ awk '
  /name=/||/age=/||/class=/||/marks=/{nr[NR]; nr[NR+1]}; NR in nr
' student.txt
name=
1st line after name
age=
1st line after age
class=
1st line after class
marks=
1st line after marks