Question

OK i am having trouble with JQuery autocomplete.

In search bar when i start typing it just shows firstname from searchfrnds.php. What i want is that it should somehow concatenate the result from searchfrnds.php in such a way that it shows firstname as well as lastname in the search result drop down that displays.

Code for searchfrnds.php:

  <?php 
   $term = trim(strip_tags($_GET['term']));

   $query = "SELECT aid,firstname, lastname, profpic,email,abtyou
    FROM artist92 WHERE firstname LIKE '%".$term."%' OR lastname LIKE '%".$term."%'";


    $result = mysqli_query ($link, $query);


    $array = array();

    while($obj = mysqli_fetch_array($result, MYSQLI_ASSOC)){
    $array['value'] = $obj['firstname'];
$array['lvalue'] = $obj['lastname'];
$array['icon'] = $obj['profpic'];
$array['aid']=$obj['aid'];
$array['abt']=$obj['abtyou'];
$array['email']=$obj['email'];

$row_set[]=$array;


       }

        echo json_encode($row_set);

      ?>

JQuery code :

  <script>

$(function() {
    function log( message ) {
        $( "<div>" ).text( message ).prependTo( "#log" );
        $( "#log" ).scrollTop( 0 );
    }

    $( "#frndsrch" ).autocomplete({
        source: "searchfrnds.php",
        minLength: 2,
        select: function( event, ui ) {
             $( "#project-icon" ).attr( "src", "" + ui.item.icon );
            $('a').attr('href',"viewartprofile.php?aid=" + ui.item.aid ); 

                 }
    });


});
</script>

Answer 1

Assuming jQueryUI autocomplete is displaying the value part of the array, just do:

$array['value'] = $obj['firstname'].' '.$obj['lastname'];

And then that should display both as required.