UNIX for Dummies Questions & Answers

^ (caret) means beginning of line. For convenience, use a different separator than slash in the substitution if your string contains a slash.

Code:

sed -e 's%^A/A/% %g' >newfile

I'm speculating you will not really want the literal asterisk there; if you do, you need to put in a backslash before it, because asterisk, too, is a regular expression operator (repetition).

In line-oriented tools, such as regular expressions, a blank line is characterized by the fact that beginning of line (caret) is adjacent to end of line (dollar sign). Maybe you want to permit optional whitespace between those too, if you want to match "graphically empty" lines containing only whitespace characters, too.

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

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 lines like /B/C and /D/E/. Is there a wild card symbol in regex that works like an * usually does?

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.

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

duh....

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

Thanks!

As a closing comment, you can run multiple commands in a single sed invocation. This should hopefully reduce the need for temporary files.

Code:

sed -n -e '/^A/d' -e '/^$/d' -e p list

This deletes all lines which match "^A", then all empty lines, then prints whatever is left.

Learning Perl is not such a bad idea, as such (unless you're the kind of person who likes single-character commands like "d" and "p" better than "delete" and "print").