jQuery UI自动完成获取有建议但不显示它们

I have a jQuery autocomplete script in my page, and if I type in the field it works on, and as requested after 3 characters, the drop containing the expected number of suggestions pops down...sort of...as there is no text in the drop down.

If I mouse over the drop down as though I were selecting one of the suggestions, a narrow bar shows a highlight for each row where a suggestion would be, but selection of one of the empty highlights empties the field. This is not surprising. I don't really understand the handling of the return data. I have seen very simple and complex code in autocomplete examples in other posts in these forums, but using the ones I thought I understood has usually broken it to the point where I don't get the drop down at all. Thus I'm unashamedly looking for someone to provide me with the code that works so I can pick it apart and work out how it works.

My basic JS code is:

    $("#tf_txt_CLIENTNAME").autocomplete({
        source: "search.php",
        minLength: 3,//search after three characters
        dataType: 'json',
        select: function(event,ui){

            //do domething

        }

It's the do something area I just can't get right. As I said, I've tried too many variations of code in that space to mention here, and all of them broke it completely.

search.php produces the following JSON output for input of 'bra':

[{"txt_CLIENTNAME":"Braedon"},{"txt_CLIENTNAME":"Bradly"}]

from the following PHP source (borrowed and modified from another web source):

<?php
require_once 'includes/dbiconnect.php';
require_once 'includes/sqlfunctions.php';

$term = trim(strip_tags(addslashes($_GET['term']))); //retrieve the search term that autocomplete sends

$qstring = "SELECT `txt_CLIENTNAME` FROM `tbl_client_details` WHERE `txt_CLIENTNAME` LIKE '%".$term."%'";
$result = mysqli_query($db_link,$qstring); //query the database for entries containing the term

while ($row = mysqli_fetch_assoc($result)) {//loop through the retrieved values
        $row['txt_CLIENTNAME']=htmlentities(stripslashes($row['txt_CLIENTNAME']));
        $row_set[] = $row; //build an array
}
echo json_encode($row_set); //format the array into json data
?>

I simply would like it to display what it finds and plonk the one I select into the value of the text field. I can then update the other fields I want to populate (company, phone # & e-mail) using $.POST triggered by the onchange event for that field (I'm sure there are better ways to do that, but I understand how to do that for the moment).

I'm using jQuery 1.7.2 and jQuery UI 1.8.22. I have the latest jquery toolbox loading as well without the tabs component, but removing it has made no difference.

Thanks in advance,

Braedon

As per Jquery UI guide,

your JSON needs to contain label or value (or both)

You need to change your json as below

while ($row = mysqli_fetch_assoc($result)) {//loop through the retrieved values
        $row['txt_CLIENTNAME']=htmlentities(stripslashes($row['txt_CLIENTNAME']));
        $row_set[] = array('label'=>'txt_CLIENTNAME','value'=>$row['txt_CLIENTNAME']); //build an array
}