Shell scripting: print first line and last line only
 

Is there a way to print the first line of a file and the last line of a file with one command?

For example:

1
2
3
4
5

and I only want it to be

1
5

Assume that I don't know how many lines are in between the first and last line. The only way I could think of was to do:

head -n 1 filename > tempfile
tail -n 1 filename >> tempfile

That works great! now, how about something a little more trickier. What if the list looked like this and I wanted to print from "/program/" to the end

apple
oranges
789
/program/
1
2
3
4
5

Assume that my list is random, but I always want to start from "/program/" until the end.

And also, would there be a way to start and stop at a certain point? For example, start at "/program/" and stop at "3"

Hi,

You can use regular expression to express line numbers in sed. I.e:

'Normal':

sed -n '15,$p' infile
Would print line 15 (included) to end of file.

Using regexp:

sed -n'/\/program\//,$p' infile
Would print from the first found /program/ to end of file.

Hope this helps.

In the case of searching from one regexp to another, you have a situation where the second regexp is a number.

You get the desired result by putting both regexp inside / like this