1. wildcard_match question
include wildcard.e
? wildcard_match("*ema*l*"," the email address")
? wildcard_match("*ema*l*","* the email address")
Is it meant to work like this? I thougth string st2 ("* the email address")
matches pattern st1 ("*em*l*") in the second case.
Best regards,
Salix
2. Re: wildcard_match question
Salix wrote:
>
>
> }}}
<eucode>
> include wildcard.e
> ? wildcard_match("*ema*l*"," the email address")
> ? wildcard_match("*ema*l*","* the email address")
> </eucode>
{{{
>
> Is it meant to work like this? I thought string st2 ("* the email address")
> matches pattern st1 ("*em*l*") in the second case.
>
I took a quick look at this. What is happening is that the * in st2 is being
treated as a literal match against the first * in st1, and hence that first * is
no longer treated as a wildcard. I noted that
? wildcard_match("*ema*l*","*email address")
prints 1 because st1 and st2 begin with the literal "*ema". I also tried:
? wildcard_match("*ema*l*"," *the email address")
which also prints 1, so obviously a * in st2 does not have to literally match
and "absorb" a * in st1.
I haven't tested this properly, but I think the following might fix it - any
comments Rob? In wildcard_match(), replace
if not find(pattern[p], {str[f], '?'}) then
with
integer pp
pp = pattern[p]
if pp='*' or (pp!=str[f] and pp!='?') then
or possibly, since the routine appears as if it would work with complex nested
sequences as well as plain character strings (and why not):
object pp
pp = pattern[p]
if equal(pp,'*') or (not equal(pp,str[f]) and not equal(pp,'?')) then
Regards,
Pete
3. Re: wildcard_match question
- Posted by Robert Craig <rds at RapidE?phoria.com>
Jul 10, 2007
-
Last edited Jul 11, 2007
Pete Lomax wrote:
>
> Salix wrote:
> >
> >
> > }}}
<eucode>
> > include wildcard.e
> > ? wildcard_match("*ema*l*"," the email address")
> > ? wildcard_match("*ema*l*","* the email address")
> > </eucode>
{{{
> >
> > Is it meant to work like this? I thought string st2 ("* the email address")
> > matches pattern st1 ("*em*l*") in the second case.
> >
> I took a quick look at this. What is happening is that the * in st2 is being
> treated as a literal match against the first * in st1, and hence that first
> * is no longer treated as a wildcard. I noted that
> }}}
<eucode>
> ? wildcard_match("*ema*l*","*email address")
> </eucode>
{{{
> prints 1 because st1 and st2 begin with the literal "*ema". I also tried:
> }}}
<eucode>
> ? wildcard_match("*ema*l*"," *the email address")
> </eucode>
{{{
> which also prints 1, so obviously a * in st2 does not have to literally match
> and "absorb" a * in st1.
>
> I haven't tested this properly, but I think the following might fix it - any
> comments Rob? In wildcard_match(), replace
> }}}
<eucode>
> if not find(pattern[p], {str[f], '?'}) then
> </eucode>
{{{
> with
> }}}
<eucode>
> integer pp
> pp = pattern[p]
> if pp='*' or (pp!=str[f] and pp!='?') then
> </eucode>
{{{
> or possibly, since the routine appears as if it would work with complex nested
> sequences as well as plain character strings (and why not):
> }}}
<eucode>
> object pp
> pp = pattern[p]
> if equal(pp,'*') or (not equal(pp,str[f]) and not equal(pp,'?')) then
> </eucode>
{{{
Thanks for looking into this.
I haven't looked at this code in 15 years.
If you think you have a good fix, feel free to check it in.
(TortoiseSVN is easier than you think to set up! 
)
Regards,
Rob Craig
Rapid Deployment Software
http://www.RapidEuphoria.com
