jQuery自动建议示例不起作用 [英] jquery auto-suggestion example doesn't work
问题描述
这对你们中的某些人来说可能很容易，但对我来说却很困难.
This might be very easy for some of you but very hard for me since first time doing it.
通过查看网络上的一些示例，我得到了下面的自动提示示例代码，但是该代码不起作用.
By looking at some examples on the web, I ended up with the code below for auto-suggestion example but the code doesn't work.
谢谢
HTML
<html>
<head>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function()
{
$('#textbox_postcode').autocomplete(
{
source: 'search-db.php',
minLength: 3
});
});
</script>
</head>
<body>
<form action="search.php" method="post">
<input type="text" id="textbox_postcode" value="" /> <input type="submit" value="Search" />
</form>
</body>
</html>
PHP
$ keyword = ltrim(strtolower(strip_tags($ _ GET ['keyword']))))
$keyword = ltrim(strtolower(strip_tags($_GET['keyword'])));
如果(！$ keyword)返回；
if (! $keyword) return;
$ host = '本地主机'; $ user ='root'; $ pswd =''; $ dtbs ='geomaps';
$host = 'localhost'; $user = 'root'; $pswd = ''; $dtbs = 'geomaps';
$ host_conn = mysql_connect($ host，$ user，$ pswd); $ dtbs_conn = mysql_select_db($ dtbs);
$host_conn = mysql_connect($host, $user, $pswd); $dtbs_conn = mysql_select_db($dtbs);
$ return = array();
$return = array();
$ sql ="SELECT id，邮政编码来自邮政编码，而邮政编码喜欢 '$ keyword％'ORDER BY邮政编码; $ run = mysql_query($ sql);
$sql = "SELECT id, postcode FROM postcodes WHERE postcode LIKE '$keyword%' ORDER BY postcode"; $run = mysql_query($sql);
如果(@mysql_num_rows($ run)== 0)返回;
if (@mysql_num_rows($run) == 0) return;
while($ records = mysql_fetch_array($ run，MYSQL_ASSOC)){$ return [] = $ records; }
while ($records = mysql_fetch_array($run, MYSQL_ASSOC)) { $return[] = $records; }
echo json_encode($ return);
echo json_encode($return);
推荐答案
尝试将$_GET['keyword']更改为$_GET['term']
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