Regex in bash?

BeholdMyGlory · 01-09-2009, 02:49 PM

Say I have a string conatining a few words:
STRING="one two three four five six seven";

how do I go about removing the first three words? I've search google but not yet found anything that seems to do what I want.

I would like to use the following regex:
"[^ ]* [^ ]* [^ ]* "
As demonstrated i Java:

Code:

String string = "one two three four five six seven";
string = string.replaceFirst("[^ ]* [^ ]* [^ ]* ","");
System.out.println(string)

I've tried
echo ${STRING/[^ ]* [^ ]* [^ ]* /}
but that just gives me the very last word.

antegallya · 01-09-2009, 04:30 PM

Hello,
it is simply because the substring replacement you are using is expanding to the longest match. (Maybe what you do isn't what you mean, the '*' in bash doesn't mean the same as the '*' in java's regexp or others. Bash's '*' means expanding to any string of any character and it doesn't expand only the preceeding pattern)
Try using this instead :

Code:

${STRING#[^ ]* [^ ]* [^ ]* }

the '#' is for deleting the following match, starting from the front of $STRING and matching the shortest match.
In java '${string#PATTERN}' would be something like

Code:

string.replaceFirst("^PATTERN","")

Regards,
antegallya

openSauce · 01-10-2009, 03:26 AM

^^ the above doesn't work for me. However you can use sed:

Code:

$ echo $STRING | sed 's/[^ ]* [^ ]* [^ ]* //'
four five six seven

Type info sed to find out how it works.

Edit: actually antegallya's works too, just had a typo before

archieval · 01-10-2009, 07:30 AM

u can also use perl, its a practical extraction =)

colucix · 01-10-2009, 08:32 AM

Or just use cut

Code:

$ string="one two three four five six seven"
$ echo $string
one two three four five six seven
$ string=$(echo $string | cut -d" " -f4-)
$ echo $string
four five six seven