regex extract substring

Question

Hi I have some SQL code that I am trying to parse and extract some text.

CREATE OR REPLACE VIEW the_view_name as

I want the text between REPLACE VIEW and the AS which does not include leading or trailing spaces.

At the moment I'm getting the whole lot with /REPLACE\s*VIEW(.*?)\s*AS/gi.

Thanks

Gilles Quénot · Accepted Answer · 2013-03-28 23:36:16Z

3

Try doing this :

 echo 'CREATE OR REPLACE VIEW the_view_name as' |
     perl -lne '/replace\s+view\s+(.*?)\s+as\b/i && print "[$1]"'
[the_view_name]

You can use this regex too in your case :

/replace\s+view\s+(\S+)\s+as\b/i

See perldoc perlrebackslash

\S : Character class for non whitespace

answered Mar 28, 2013 at 23:36

Gilles Quénot

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1 Comment

jaybee Over a year ago

$sqlline =~ /replace\s+view\s+(.*?)\s+as\b/i; $viewname = $1; print "viewname : $viewname\n";

lxop · Accepted Answer · 2013-03-28 23:36:16Z

0

You're currently missing a space after VIEW, and if you want everything except spaces then your capture group should be ([^\s]*), which will collect everything up until a space.

answered Mar 28, 2013 at 23:36

lxop

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1 Comment

jaybee Over a year ago

Thanks;REPLACE\sVIEW\s*([^\s]) gives me REPLACE VIEW the_view_name I'm still after only 'the_view_name'

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