How to do multiple grep pattern to find value in grepped string - regex

Question

I am trying to do multiple grep pattern to find a number within a grepped string.
I have a text file like this:
This is the first sample line 1
this is the second sample line
another line
total lines: 3 tot
I am trying to find a way to get just the number of total lines. So the output here should be "3"
Here are the things I've tried:
grep "total lines: [0-9]" myfile.txt
grep "total lines" myfile.txt | grep "[0-9]"

Answer 1

You could use sed:
sed -En 's/^total lines: ([0-9]+).*/\1/p' myfile.txt
-E extended regular expressions
-n suppress automatic printing
Match ^total lines: ([0-9]+).* (capture the number)
\1 replace the whole line with the captured number
p print the result

Answer 2

1st solution: Using GNU grep try following. Simply using -o option to print only matched value, -P enables PCRE regex for program. Then in regex portion matching from starting ^total lines: in each line and if a match found then discard matched values by \K option(to remove it from expected output) which is followed by 1 or more digits, using positive look ahead to make sure its followed by space(s) tot here.
grep -oP '^total lines: \K[0-9]+(?=\s+tot)' Input_file
2nd solution: With your shown samples, please try following in awk. This could be done in a single awk itself. Searching line which has string /total lines: / in it then printing 2nd last field of that line.
awk '/total lines: /{print $(NF-1)}' Input_file
3rd solution: Using awk's match function here. Matching total lines: [0-9]+ tot and then substituting everything apart from digits with null in matched values.
awk 'match($0,/total lines: [0-9]+ tot/){val=substr($0,RSTART,RLENGTH);gsub(/[^0-9]+/,"",val);print val}' Input_file

Answer 3

Do you have to use grep?
$ echo myfile.txt | wc -l
If you mean that the file has a line in it formatted as
total lines: 3 tot
Then refer to https://unix.stackexchange.com/questions/13466/can-grep-output-only-specified-groupings-that-match and use something like:
grep -Po 'total lines: \K\d+' myfile.txt
Notes:
Perl regex is not my forte, so the \d\w part might not work.
This may be doable without -P, but I cannot test from this windows computer.
regex101.com helped me test the above line, so it may work.

Answer 4

Problem with relying on pattern of last line and applying grep/sed to find pattern is that if any line in file contains such pattern, then you will have to apply some additional logic to filter that.
e.g. Consider case of below input file.
line001
total lines: 883 tot
This is the first sample line 1
this is the second sample line
another line
total lines: 883 tot
Assuming your file format is constant (i.e. Second last line will be blank and last line will contain total count), instead of using any pattern matching commands you can directly count number of rows using below awk command.
awk 'END { print NR - 2 }' myfile.txt

Answer 5

You can use the following awk to get the third field on a line that starts with total count: and stop processing the file further:
awk '/^total lines:/{print $3; exit}' file
See this online demo.
You can use the following GNU grep:
# Extract a non-whitespace chunk after a certain pattern
grep -oP '^total lines:\s*\K\S+' file
# Extract a number after a pattern
grep -oP '^total lines:\s*\K\d+(?:\.\d+)?' file
See an online demo. Details:
^ - start of string
total lines: - a literal string
\s* - any zero or more whitespace chars
\K - match reset operator discarding all text matched so far
\S+ - one or more non-whitespace chars
\d+(?:\.\d+)? - one or more digits and then an optional sequence of . and one or more digits.
See the regex demo.