From Regex to Regex

Question

sed -n '/regexp/,$p'
Was trying just having it twice..and a few other attempts but really none that worked

Answer 1

I cant get that to work, and havent heard too much about IFS..let me start over real fast..
I have a file that i broke up into two parts.
These Commands:
cat myfile.txt | sed -n '/regex/,/regex/{/regex/d;p}' | sed -n '/---/,+3!p' | sed -n '/#/!p' | sed 's/^[ ]*//' | sed -e s/[^:]*://
Gives me the first part of the file.
And These Commands:
cat myFile.txt | sed -n '/regex/,/regex/{/regex/d;p}' | sed -n '/#/!p' | sed -e s/[^:]*:// | sed /--------/,+10d
Basically spliting the file up into two parts.
Now..Finally...I can redirect to a text file. So do a chmod +X myfile.txt so i can execute it.
But wait... I need to execute them one by one..and compare the output..
So yeah, scripting this would be the way to go..Your right i would have two arguments. But when i use my.txt my2.txt nothing happens.
Again i need to execute each line of commands one by one.. The File looks like this:
find /home -name ".bashrc" -printf "%h\n" 2>/dev/null | sed -r 's/.*\/([^\/]*$)/\1/'
grep -e "^[^:\.]*:" /etc/passwd | cut -d':' -f1
grep -E "^[^:]{11,}:" /etc/passwd | cut -d':' -f1
find / -regextype posix-extended -regex ".*\/[^\/]{11,}$" -printf "%H%f\n"2>/dev/null
And FileB:
find / -type f -name '.bashrc' -print -exec grep PS1 '{}' \; 2>/dev/null | xargs ls -l 2>/dev/null
find /home -name '.bashrc' 2>/dev/null
cat /etc/passwd | cut -d':' -f1 | grep -v "\." | sort
cat /home/student/joseph.hodges/item4.txt | cut -d"/" -f4 | egrep '[A-Za-z]{10,}$'
cat /etc/passwd | cut -d':' -f1 | egrep '[A-Za-z]{10,}$'
du / -b 2>/dev/null | egrep '^.*/[^/]{10,}$' | head -20
I've been trying to figure out how i can execute a line from a each file upon specification of $1 and $2 and then compare them.
Last edited by oxoxo (2008-12-03 06:03:45)