Попробуйте (выборка базы данных с несколькими полями mydb = data)
//demo_cities.php
<?php
$con=mysqli_connect("localhost","root","","auto");
$return_arr = array();
$ac_term = "%".$_GET['term']."%";
$query = "SELECT * FROM data where name like '$ac_term'";
$result=mysqli_query($con,$query);
while ($row = mysqli_fetch_row($result))
{
$row_array['label'] = $row[0];
$row_array['price'] = $row[1];
array_push($return_arr,$row_array);
}
echo json_encode($return_arr);
?>
//index.php
</head>
<body>
<form method="post">
<fieldset>
<p><label>Name: </label>
<input type="text" id="name" name="name" /> <br />
<label>Price: </label>
<input type="text" id="price" name="price" /> <br /></p>
</fieldset>
</form>
<script>
$(function() {
$('#price').val("");
$("#name").autocomplete({
source: "demo_cities.php",
minLength: 1,
select: function(event, ui) {
$('#price').val(ui.item.price);
},
response: function( event, ui )
{
$('#price').val("");
}
});
});
</script>
</body>
</html>