Egrep - find 0 (zero) and ignore previous line

Question

I am trying hard to get the output as I Like.

Current Output:

###Server1###
2
###Server2###
0
###Server3###
     5
###Server4###
     0

Required Output:

###Server1###
2
###Server3###
     5

All I am looking is to grep and ignore any line and the previous line that containts 0 (zero) in any place of the line. I am using bash shell.

Answer 1

This is a possible approach:

$ grep -B 1 "^\s*[1-9]$" file
###Server1###
2
--
###Server3###
     5

To get rid of the group separator, we can also do:

$ grep --no-group-separator -B 1 "^\s*[1-9]$" file
###Server1###
2
###Server3###
     5

Explanation

Instead of using grep -v to find the inverse, I think it is easier to look for the lines having a single digit value not being 0. This is done with the "^\s*[1-9]$" expression, that allows spaces before the digit.

With -B 1 we make it print also the line before the matched one.

Answer 2

Code for GNU sed:

sed '$!N;/\s*\b0\b\s*/d' file

$ sed '$!N;/\s*\b0\b\s*/d' file
###Server1###
2
###Server3###
     5