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Thanks again, you've really been helpful and I am finally done with this project.

Incidentally, this article was really helpful:

HANDY ONE-LINERS FOR SED...

duh....

the answer was to run the following:
sed -n '/^A/!p' list > list2
sed '/^$/d' list2 > list4

Thanks!

sed -n '/^A/p' list > file
selects all the lines I don't want. Can I negate it and select all the others?

alternatively, I could select all lines that DON'T begin with that string but I'm not sure how to do that yet. I'm trying to figure it out, but any help would be appreciated.

Ok... so this deletes all instances of "A/A/" but I actually want to cut any lines that being with "A/A/". Many of these are in the format A/A/B/C and A/A/D/E and this leaves me with a bunch of...

Thank you very much for all your help and for not telling me to learn perl :)

sigh... still working on this.

There are 2 more things I have to figure out. .

How do you indicate all lines beginning with specific file paths for example:

A/A/*

sed -e 's/"A/A/*"/...

got it, thanks.

The answer was to use cut as such:

cut --delimiter='%' -f 1 list3 > name
cut --delimiter='@' -f 2 list3 > location

Thanks for the help!

I changed the command to read
sed -e 's/ - / % /' list > list3

and now I have a separate symbol to break up the data. Is there an expression to say cut all data up to the % and write to a file?

well that removed all the hyphens between name and location WITHOUT removing the words that needed to be hyphenated, so that's a start, but it still doesn't allow me to separate the data on either...

Hi all,

I'm a beginner with linux, regex, grep, etc
I am trying to get data out of a file that has about 13,000 lines in this format

name - location

I want to grep all the names out to one...