Shell Programming and Scripting

Say I have a file 'example.txt' with these lines of code in it:

hello:anddasd:cheese:gerg
whatever:sdadsa:asdfasdfa:wwew
hmmmm:something:gfhfhgf:sdasdas

Question:
1. How would I write a script which is able to take all the words before the first ':'?
2. How would I write a script which is able to take all the words between the last two ':'s e.g. with "hello:anddasd:cheese:gerg" I'd be able to get the word "cheese"?

I've been looking at tutorials but they mostly focus on using replace so I'm stuck. Any ideas?

Thanks

try this:

Code:

awk 'BEGIN { FS=":" } { print $1 $2 $3 }' try

Sorry, I'm unfamiliar using awk and I really want to learn how to do this with sed. So how would I go about doing this in sed?

Thanks

Code:

cut -d":" -f2- file

Code:

cut -d":" -f3 file

Code:

sed "s/[^:]*://" file

Code:

sed "s/:[^:]*$//;s/.*://" file

Jesus that was fast. I was spending hours on this yesterday trying to figure it out and you did it in a minute. So let me just go through this so I understand:

sed "s/:[^:]*$//;s/.*://" file

so 's/' puts it in select mode

1.
:[^:]*$ here's the confusing bit. So what does this mean? Here's me guessing.
Choose ':' and any character at the beginning of ':' when there are 0 or more occurrences of the preceding character before the end of the line??? I'm sure what I said is wrong but that's a guess.

2.
//; keep doing this on each line till the end of the file?

3.
s/.*:// I think I get this bit. This means any character before the next ':' right?

Help out a noob, please.
Oh, and the first line you gave me gives up to the third ':'. I just want the words before the first ':' for that bit.

Please check man SED.

Code:

sed "s/:.*//" file