Thank you all.
On Thu, Feb 13, 2014 at 10:47 PM, Walter Prins <wpr...@gmail.com> wrote:
> Hi,
>
> On 13 February 2014 06:44, Santosh Kumar <rhce....@gmail.com> wrote:
> > I am using ipython.
> >
> > 1 ) Defined a string.
> >
> > In [88]: print string
> > foo foobar
> >
> > 2) compiled the string to grab the "foo" word.
> >
> > In [89]: reg = re.compile("foo",re.IGNORECASE)
> >
> > 3) Now i am trying to match .
> >
> > In [90]: match = reg.match(string)
> >
> > 4) Now i print it.
> >
> > In [93]: print match.group()
> > foo
> >
> > Correct me if i am wrong, i am expecting both "foo" and "foobar", why is
> it
> > giving
> > just "foo"
>
> A small addition to Peter's already comprehensive reply: Your regular
> expression is not including what follows "foo", it is defined as
> *only* the string literal "foo", so it can only ever match and return
> the literal string "foo".
>
> Try specifying "foo.*" as the regular expression. Example session:
>
> Python 2.7.5 (default, May 15 2013, 22:43:36) [MSC v.1500 32 bit (Intel)]
> Type "copyright", "credits" or "license" for more information.
>
> IPython 1.0.0 -- An enhanced Interactive Python.
> ? -> Introduction and overview of IPython's features.
> %quickref -> Quick reference.
> help -> Python's own help system.
> object? -> Details about 'object', use 'object??' for extra details.
>
> [C:/Src]|1> s='foo foobar'
>
> [C:/Src]|2> import re
>
> [C:/Src]|3> reg=re.compile('foo.*', re.IGNORECASE)
>
> [C:/Src]|4> match=reg.match(s)
>
> [C:/Src]|5> print match.group()
> foo foobar
>
>
>
> Walter
>
--
D. Santosh Kumar
RHCE | SCSA
+91-9703206361
Every task has a unpleasant side .. But you must focus on the end result
you are producing.
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