UNIX for Dummies Questions & Answers

can someone please explain the below sed command..

Code:

 
sed 's/\([^|][^|]*|\)\(.*\)/\2\1/'

Code:

[house@leonov] echo "a|b|c|d|e" | sed 's/\([^|][^|]*|\)\(.*\)/\2|\1/'
b|c|d|e|a|

i know what the sed command does, just do not know how it does that...... so was hoping if someone could explain it......

It looks for at least one character that is not a '|' ( [^|] ) at the start of the line plus the first '|' and stores it in variable 1 it stores the rest of the line in variable 2. Then it prints variable 2 followed by variable 1 ( \2\1 ). IMO in practice it is equivalent to:

Code:

sed 's/\([^|]*|\)\(.*\)/\2\1/'

In short, what this will do is the following: Given an arbitrary string containing one or more characters that are not | followed by a |, take everything after that | character and move it to the front of that whole pattern. So, for example, it would turn 'foo|bar' into 'barfoo|', but '||foo|bar' would become '||barfoo|'. The first | is bypassed because it has nothing on its left; the second is skipped because it has only another | on its left. 'bar' is moved after the leading || characters because they are not part of the pattern matched.

(Whether this is what was actually intended is another question.)

As a side note, 's/\([^|][^|]*|\)\(.*\)/\2\1/' is equivalent to the simpler pattern 's/\([^|]+|\)\(.*\)/\2\1/'. [^|] is the set of all characters which are not |, so [^|]* matches any zero or more characters that are not |, while [^|]+ matches any one or more such characters, which makes [^|][^|]* functionally identical to [^|]+ within the regexp.

O'Reilly has a good book on regular expressions. It covers things like this and much more.

Not all sed-s support 'pattern+' paradigm - Solaris' sed doesn't.

Corrrect, AFAIK Only GNU sed supports pattern+ as pattern\+ .
+ is not part of POSIX BRE (Basic Regular Expressions)