我一直在搞這個太久,試圖讓它工作。任何人都可以請看看你是否有任何指針。與PHP的Jquery自動完成功能
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<link type="text/css" rel="stylesheet" href="autocomplete.css" />
<script src="jquery-1.7.2.min.js" type="text/javascript"></script>
<script src="jquery-ui-1.8.21.custom.min.js" type="text/javascript"></script>
<title></title>
</head>
<body>
<style>
.ui-autocomplete-loading { background: white url('images/ui-anim_basic_16x16.gif') right center no-repeat; }
</style>
<script>
$(function() {
$("#materials").autocomplete({
source: "autocomplete.php",
minLength: 2
});
});
</script>
<div class="demo">
<div class="ui-widget">
<label for="materials">Materials: </label>
<input id="materials" />
</div>
</div><!-- End demo -->
</body>
</html>
和PHP文件是
require_once "db_con.php"; // Database connection, I know this works.
$q = strtolower($_GET["q"]);
if (!$q)
return;
$sql = "SELECT * FROM materials WHERE name LIKE '%$q%'";
$rsd = mysqli_query($dbc, $sql) or die(mysqli_error($dbc));
while ($rs = mysqli_fetch_array($rsd)) {
$cname = $rs['name']; // I know this all returns correctly
echo json_encode($cname); // First time I have ever used json, error might be here.
}
我想有動力的jQuery自動完成是從MySQL使用PHP提供的數據的網頁。 Simples。只有它不工作...
任何人有任何想法我失蹤?
問候
---- ----編輯
爲了確認這是工作,我完成以下任務:
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<link type="text/css" rel="stylesheet" href="autocomplete.css" />
<script src="jquery-1.7.2.min.js" type="text/javascript"></script>
<script src="jquery-ui-1.8.21.custom.min.js" type="text/javascript"></script>
<title></title>
</head>
<body>
<style>
.ui-autocomplete-loading { background: white url('images/ui-anim_basic_16x16.gif') right center no-repeat; }
</style>
<script>
$(function() {
$("#materials").autocomplete({
source: <?php
include_once 'db_con.php';
$sql = "SELECT name FROM materials";
$rsd = mysqli_query($dbc, $sql) or die(mysqli_error($dbc));
echo '[';
while ($rs = mysqli_fetch_array($rsd)) {
echo "'" . $rs['name'] . "', "; //add results to array
}
echo ']';
?>,
minLength: 2
});
});
</script>
<div class="demo">
<div class="ui-widget">
<label for="materials">Materials: </label>
<input id="materials" />
</div>
</div><!-- End demo -->
</body>
</html>
這完美的作品。那麼好INFACT我想我要保持這種代碼並沒有它應該如何工作的,但...
編輯是構建JSON字符串最可怕的方式。嘗試:'$ json = array(); while($ rs = mysqli_fetch_array($ rsd)){$ json [] = $ rs ['name']; }回聲json_encode($ json);' – Sammaye 2012-07-19 09:23:56