jQuery的自动完成不返回建议数据onselec

Question

I have input with id aktForm_tiekejas and is jquery autocomplete code:

$('#aktForm_tiekejas').autocomplete({
serviceUrl: '_tiekejas.php',
width: 185,
deferRequestBy: 0,
noCache: true,
onSelect: function(suggestion) {alert('You selected:'+suggestion.value+','+suggestion.data);}
});

_tiekejas.php:

<?php
include("../Setup.php");
$query = ($_GET['query']);
$reply = array();
$reply['query'] = $query;
$reply['suggestions'] = array();
$reply['data'] = array();
$res = mysql_query("SELECT id,pavadinimas FROM sarasas_tiekejas WHERE pavadinimas LIKE '%$query%' ORDER BY pavadinimas ASC");
while ($row = mysql_fetch_array($res)) {
 $reply['suggestions'][] = $row['pavadinimas'];
 $reply['data'][] = $row['id'];
}
mysql_close();
echo json_encode($reply);
?>

If query is 'vac' php returns from server:

{"query":"vac","suggestions":["UAB Vivacitas"],"data":["866"]}

but

alert('You selected:'+suggestion.value+','+suggestion.data); 

doesn't alert data (866)

why?...

Answer 1

Probably because suggestion.value doesn't exist. Looking at your JSON response code I can see the suggestion.data but no suggestion.value. Since JS will be looking for a value that doesn't exist it will throw an error. Also you're missing out the array for the second part of the return. Try this:

alert('You selected: '+suggestion.data[0]); 

If you need to iterate through your data subsets do something like:

for(i in suggestion.data){
    alert('You selected: '+suggestion.data[i]);
}