jQuery UI自动完成ajax未填充下拉框 [英] Jquery UI autocomplete ajax is not populating dropdown box
问题描述
我需要帮助，我看不出问题出在哪里.
I need help with this, I can't see where is the problem.
当我在html文件中设置自动完成源时，它工作正常，当我在ajax.php中打印出相同的源或数据库值并通过ajax返回时，它不起作用.可能是什么问题呢?请帮助.
When I set source for autocomplete in html file, it works fine, when I same source or database values print out in ajax.php and return it via ajax it doesn't work. What could be the problem? Help please.
HTML:
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Auto complete</title>
<script type="text/javascript" src="jquery-1.7.1.min.js"></script>
<script type="text/javascript" src="jquery-ui-1.8.18.custom.min.js"></script>
<link rel="stylesheet" media="all" type="text/css" href="jquery-ui-1.8.custom.css" />
<style type="text/css">
.ui-autocomplete-loading {
background: url("images/loader.gif") no-repeat scroll right center white;
}
</style>
<script type="text/javascript">
jQuery(document).ready(function($){
$("#ac").autocomplete({
minLength: 2,
//source: [{"value":"Some Name","id":1},{"value":"Some Othername","id":2}]
source: function( request, response){
$.ajax({
type: 'GET',
url: 'ajax.php',
data: {
'term':request.term
},
contentType: "application/json; charset=utf-8",
dataType: "json",
success: function(data){
console.log('Success : ' + data);
},
error: function(message){
alert(message);
}
});
},
select: function( event, ui ) {
}
});
});
</script>
</head>
<body>
<input type="text" id="ac" name="ac" size="100" />
</body>
</html>
和我的ajax.php文件:
and my ajax.php file:
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$dbname = 'test_db';
$server = mysql_connect($dbhost, $dbuser, $dbpass);
$connection = mysql_select_db($dbname, $server);
$term = $_GET['term'];
$qstring = "SELECT user_id,tName FROM `csv_data` WHERE tName LIKE '%" . $term . "%'";
$result = mysql_query($qstring);
$return_arr = array();
$i = 0;
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {//loop through the retrieved values
$row_a = array();
if ($row['tName'] != null) {
$i++;
$row_a['value'] = ($row['tName']);
$row_a['id'] = (int) $i;
array_push($return_arr, $row_a);
}
}
mysql_close($server);
header("Content-type: text/x-json");
/*$my_arr = array(
array("value" => "Some Name", "id" => 1),
array("value" => "Some Othername", "id" => 2)
);*/
//echo json_encode($return_arr);
print json_encode($return_arr);
//print json_encode($my_arr);
这是萤火虫(从数据库生成)的响应.
This is response from firebug(generated from database).
[{"value":"4 erste Spiele","id":1},{"value":"Meine ersten Spiele \"Blinde Kuh\"","id":2},{"value":"4 erste Spiele","id":3},{"value":"Meine ersten Spiele \"Blinde Kuh\"","id":4},{"value":"4 erste Spiele","id":5},{"value":"Meine ersten Spiele \"Blinde Kuh\"","id":6},{"value":"Maxi Kleine Spielewelt","id":7}]
推荐答案
参数response实际上是一个回调，您必须调用该回调-将数据传递给它-以显示结果弹出菜单.只需在成功"回调中调用它即可:
The parameter response is actually a callback tthat you have to call - passing it your data - to display the result popup menu. Simply call it in the "success" callback:
source: function(request, response) {
$.ajax({
...
success: function(data) {
// pass your data to the response callback
response(data);
},
error: function(message) {
// pass an empty array to close the menu if it was initially opened
response([]);
}
});
},
这篇关于jQuery UI自动完成ajax未填充下拉框的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
