Conditionally delete last X lines

Shell Programming and Scripting
1

delete last X lines,which start with [0-9]+

example file:

test1
test2
[23] remove1 
[24] remove2

one liner shell is preferred.

2

so what did you try?

3

Also, I want a cup of coffee and a croissant!

Why don't you tell us what you have tried so far?

4

I figured it out, But I don't like the trick of tac.
[Help still needed, Solaris doesn't have tac command.]

$cat t2
test1
test2
[23] remove1 
test3
[24] remove2
[1] remove

$tac t2 | awk '{ if (($0 ~ /^\[[0-9]+\]/ )&&( flag!=1 )) { } else {print $0; flag=1 } } ' | tac
test1
test2
[23] remove1 
test3
5

from above sample, what do you need removed? All the lines within [ ] ?

test1
test2
[23] remove1 
test3
[24] remove2
[1] remove
6

Could this help you.

awk '{a[++i]=$0} END{for(j=i;j>=1;j--){print a[j]}}' inputfile | awk '{if (($0 ~ /^\[[0-9]+\]/ )&&( flag!=1 )) { } else {a[++i]=$0;flag=1 } } END{for(j=i;j>=1;j--){print a[j]}}'
7 
awk '/^\[[0-9]+\]/{s=s (s""?RS:x) $0;next}s""{print s; s=x}1' infile
8

Hi Scrutinizer , Could you please explain the code ?

9

Hi Pravin, if the line starts with one or more numbers in square brackets then append the line to string s. If the string is not empty then insert a newline character in between and process the next line..
Else (if the line start with something else), if the string s is not empty then print it and set it to an empty string. Print the current line (1).

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