Hi Daniel,
I'll answer in English.
On 27.06.02 (11:43), Daniel Hegermann wrote:
> Ich will alle Mitglieder aus einer bestimmen Mitglieder Gruppe (Form)
> auslesen (Gruppen ID 5 (laut NAdmin)).
>
> Aber es werden immer nur die NAdmin Administation Mitglieder angezeigt. Wenn
> ich eine andere Gruppe ausw�hle, dann erhalte ich auch nur die NAdmin
> Mitglieder. Bei einer ID, die nicht existiert gibt es eine Object
> Fehlermeldung.
>
> Ist der Befehl richtig um eine Gruppe auszulesen.
>
> English:
>
> I want to determine all members from one members group (form) to select
> (groups of ID 5 (according to NAdmin)).
>
> But in each case the NAdmin Administation of members is indicated. If I
> select another group, then I receive also only the NAdmin of members. With an
> ID, those not existed gives it a Object to error message.
>
> The instruction is to be picked out correctly around a group.
>
>
> Script:
>
> Line 72: $group=5;
> Line 73: $lst = mgd_list_members($group);
> Line 74: if($lst){
> Line 75: while( $lst->fetch() ) {
> Line 76: $person = mgd_get_person($lst->uid);
> Line 77: if($person->username==$usersid and
> $person->password==$passsid){
> Line 78: ..
> Line 79: }
> Line 80: }else{
> Line 81: die("Keine Verbindung");
> Line 82: }
Generally, mgd_list_members() is the right function for the
task. If I understand you right (from the German part of your
e-mail), you always get a list of members of the Admin Group,
regardless of the id supplied to mgd_list_members()?
This is ... should be ... impossible, of course ;-)
Could you verify this behaviour by displaying lists of members
and comparing those by hand? Your code:
$group = 5; // Change ID and look whether the list created by
// the following code changes as well
$lst = mgd_list_members($group);
if ($lst) while ($lst->fetch()) echo "$lst->uid<br>\n";
else echo "Group does not exist or No members in group<br>\n";
Please make sure that your bug description is really consistent
and reproducable!
Thanks,
phr
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