Extract line number of a file which is having a non zero value before a specified string

Question

I'm having a file which contains following data

1. verification: 10 passed 0 failed
2. verification: 10 passed 0 failed
3. verification: 10 passed 1 failed
4. verification: 10 passed 3 failed
5. verification: 10 passed 0 failed

I want to know the line numbers of 3 and 4.

Answer 1

Assuming the file is called in.txt, then:-

awk '$5>0 {print $0; }' in.txt

will give you:-

3. verification: 10 passed 1 failed
4. verification: 10 passed 3 failed

I've made the assumption the the line numbers (1-5) is part of the file. If it isn't change the $5 to $4.

If the line number is part of the file and you want to just print it without the dot:-

awk '$5>0 {sub(/\./,"",$1); print $1 }' in.txt

which gives:-

3
4

Answer 2

With GNU grep, you can use lookbehind:

$ grep -P '(?<!0) failed' file 
3. verification: 10 passed 1 failed
4. verification: 10 passed 3 failed

Or a shorter version of awk:

$ awk '$5 > 0' file

Answer 3

Please note that i'm not printint real numbers, but the first number in the line until the dot.

With sed:

 cat inputfile.txt | sed  '/ 0 failed$/d ; s/\..*//'

With grep and sed:

 cat inputfile.txt | grep -v ' 0 failed$' | sed 's/\..*//'

Output:

3
4