Question

I having problems in extracting the word from a line. What i want is that it picks the first word before the symbol # but after the /. Which is the only delimiter that stand out.

A line looks like this:

,["https://picasaweb.google.com/111560558537332305125/Programming#5743548966953176786",1,["https://lh6.googleusercontent.com/-Is8rb8G1sb8/T7UvWtVOTtI/AAAAAAAAG68/Cht3FzfHXNc/s0-d/Geek.jpg",1920,1200]

I want the word Programming.

To get that line i am using this which narrows it down.

sed -n '/.*picasa.*.jpg/p' 5743548866439293105

So i want it to pretty much find # and then go backward until it hit the first /. Then print it out. In this case the word should be Programming but could be anything.

I want it to be as short as possible and have experimented with

sed -n '/.*picasa.*.jpg/p' 5743548866439293105 | awk '$0=$2' FS="/" RS="[$#]"

Answer 1

You can do that with sed (slightly shortened for formatting but works on your original string as well):

pax> echo ',["https://p.g.com/111/Prog#574' | sed 's/^[^#]*\/\([^#]*\)#.*$/\1/'
Prog
pax> 

Explaining in more detail:

    /---+------------------> greedy capture up to '/'.
   /    |
   |    | /------+---------> capture the stuff between '/' and '#'.
   |    |/       |
   |    ||       | /-+-----> everything from '#' to end of line.
   |    ||       |/  |
   |    ||       ||  |
's/^[^#]*\/\([^#]*\)#.*$/\1/'
                      ||
                      \+---> replace with captured group.

It basically searches for an entire line that has the pattern you want (first # following a /), whilst capturing (with the \( and \) brackets) just the stuff between / and #.

The substitution then replaces the entire line with just that captured text you’re interested in (via \1).