How to grep only two words in a line in file between them specific number of random words present

Question

Given a file with this content:

Feb 1 ohio a1 rambo
Feb 1 ny   a1 sandy
Feb 1 dc   a2 rambo
Feb 2 alpht a1 jazzy

I only want the count of those lines containing Feb 1 and rambo .

Answer 1

You can use awk to do this more efficiently:

$ awk '/Feb 1/ && /rambo/' file
Feb 1 ohio a1 rambo
Feb 1 dc   a2 rambo

To count matches:

$ awk '/Feb 1/ && /rambo/ {sum++} END{print sum}' file
2

Explanation

• awk '/Feb 1/ && /rambo/' is saying: match all lines in which both Feb 1 and rambo are matched. When this evaluates to True, awk performs its default behaviour: print the line.

• awk '/Feb 1/ && /rambo/ {sum++} END{print sum}' does the same, only that instead of printing the line, increments the var sum . When the file has been fully scanned, it enters in the END block, where it prints the value of the var sum .

Answer 2

Try this as per @Marc's suggestions,

grep 'Feb 1.*rambo' file |wc -l

In case, position of both strings are not sure to be as mentioned in question following command will be useful,

grep 'rambo' file|grep 'Feb 1'|wc -l

The output will be,

2

Here is what I tried,

Answer 3

Is Feb 1 always before rambo? if yes:

grep -c "Feb 1 .* rambo"

Answer 4

awk解决方案可能更清晰，但这是一种不错的sed技术：

 sed -n '/Feb 1/{/rambo/p; }' | wc -l