Re: Sorting output to be returned to a HTML page using PHP (SUCCESS)

["Antonin Novak" <noc@cca.cz>]

Doug,

Firstly I must apologise for the typos in my last message ie. the 
missing quotes and wrong format of ldap_get_entries. they were 
correct in my code.

The problem I had was that I had specified a wrong sortfield, 
instead of specifyinh "sn" as per your instructions, I used $info[sn].

I now produce sorted information.

Thanks for all your help. I think/hope  I am a little wiser now.

Tony

> On 24 May, Antonin Novak wrote:
> > Sorry if I am thick, but where does the sort write the data to?
> 
> It is sorted in place (the result is in the $info array).
> 
> > I have been trying to get the script you sent me to work, with no
> > joy. I have 107 entries in my selection, but when I sort them my
> > $info ends up with no records.
> > 
> > I have printed the $a[sortfield][0] and the $b[sortfield][0] in the
> > various "if statements" and get nothing.
> should have been $a[$sortfield][0] (php won't complain about this)
> 
> > When I list $info after the sort I get 0=107 and 1 to 107 = array.
> 
> This is correct. Because of the is_array tests, the "count" element is
> sorted to the top of the list ($info[0] = 107). Then, each successive
> element is an array of attributes. Each attribute is an array of
> values. When you print an array in php it will output "Array".
> 
> These are the same mistakes as before, I'm not sure where they are
> from but I'll point them out this time.
> 
> >                        	$justthese = array ("sn", "givenname); 
> Missing a quote
> >                        	$info=ldap_get-entries($ds,$sr); 
> Should be ldap_get_entries
> >  	                  	usort($info,EntryCompare); 
> after the sort, element 0 is the old "count" parameter. You cannot
> reference it by ["count"] since a usort remove the association. YOu
> could us a uasort but the would also keep the association of each
> value, which means you could not access each element in sorted order
> via an index. You could by using the each or next functions. >
> 					while (list($key,$value)=each ($info)) > 							{ echo "$key:
> $value\n";} Here you are printing the array itself, not the values. If
> you want a complete dump of information, you need to do something like
> 
>   while (list($key,$value)=each ($info))
>   {
>     if (!is_array($value))
>       echo "$key: $value<br>\n";
>     else
>     {
>       echo "$key:<br>\n";
>       while (list($attribute_key, $attribute_value) = each($value)) {
>         if (!is_array($attribute_value))
>           echo "- $attribute_key: $attribute_value<br>\n";
>         else
>         {
>           echo "- $attribute_key:<br>\n";
>           while (list($value_key, $value_value) =
>           each($attribute_value)) {
>             echo "- - $value_key: $value_value<br>\n";
>           }
>         }
>       }
>     }
>     echo "<br>\n";
>   }
> 
> > 			print "no of records in the sorted file =".$info["count"];
> You can't use ["count"] anymore. Use [0].
> >                        for ($i=0, $i<$info["count"]; $i++) 
> >       	                 { 
> >            		            echo statement 
> >                        	} 
> 
> I don't know where statement comes from. What I had was
> 
>   // start at 1 to skip count
>   for ($i=1; $i<count($info); $i++)
>   {
>     print($info[$i]["dn"] . "<br>\n");
>     print("-- " . $info[$i]["uid"][0] . "<br>\n");
>   }
> 
> -- 
> Doug Nazar
> Dragon Computer Consultants Inc.
> Tel: (416) 708-1578     Fax: (416) 708-8081



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  Antonin Ch. Novak                    mailto:Antonin.Novak@cca.cz
  Systemovy administrator
  Systemova a technicka podpora
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