Posted by Vedanta Barooah on 04/28/05 16:43

Well that was simple, but this is what i am trying to solve:

if you refer to the php documentation for ldap_open() function it says:

resource ldap_search ( resource link_identifier, string base_dn,
string filter [, array attributes [, int attrsonly [, int sizelimit [,
int timelimit [, int deref]]]]])

if you look at the 4th and the 6th arguments to the function
attributes is an array while sizelimit is an int, i want to pass the
sixth element without passing the 4th and the 5th ... how do i do
that??

i tried these options:

# this does not work,
$rs=ldap_search($con,"o=vodoo.com","(objectClass*)",array(),0,500);

# this wont works :((
$rs=ldap_search($con,"o=vodoo.com","(objectClass*)",' ',0,500);

# this also goofs!
$rs=ldap_search($con,"o=vodoo.com","(objectClass*)",NULL,0,500);

here that 5th arg works if i pass a zero as ... 0 means the default behaviour!!

any ideas ... clues ?

Thanks,
Vedanta Barooah




On 4/28/05, Marek Kilimajer <lists@kilimajer.net> wrote:
> Bostjan Skufca @ domenca.com wrote:
> > function add ($a=1, $b=2, $c=3) {
> > return $a + $b + $c;
> > }
> > add(1, null, 1);
> >
> > will do just fine
>
> returns 2, OP wants 4 IMO
>
>
> >
> > r.,
> > Bostjan
> >
> >
> >
> > On Thursday 28 April 2005 14:16, Marek Kilimajer wrote:
> >
> >>Vedanta Barooah wrote:
> >>
> >>>Hello All,
> >>>Cosider this :
> >>>
> >>>function add($a=1,$b=2,$c=3){
> >>> return $a + $b + $c;
> >>>}
> >>>
> >>>how do i skip the second argument while calling the function, is there
> >>>a process like this:
> >>>
> >>>echo add(1,,1); # where i expect 2 to be printed,
> >>
> >>php does not support this. you can workaround this using:
> >>
> >>function add($a = null,$b = null, $c = null){
> >> if(is_null($a)) $a = 1;
> >> if(is_null($b)) $b = 2;
> >> if(is_null($c)) $c = 3;
> >> return $a + $b + $c;
> >>}
> >>
> >>add(1, null, 1);
> >
> >
>
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>
>


--
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Vedanta Barooah
YM! - vedanta2006
Skype - vedanta2006

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