为了让它正常工作,我已经花了很长时间来处理它。有没有人能帮我看看你们有什么建议。
<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>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.
}我试图有一个由Jquery自动完成的网页,这是从mysql使用PHP提供的数据。简单。只是它不工作..。
有人知道我错过了什么吗?
问候
--编辑
为了检查它是否正常工作,我完成了以下操作:
<!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 "'" . $rs['name'] . "', "; //add results to array
}
echo ']';
?>,
minLength: 2
});
});
</script>
<div class="demo">
<div class="ui-widget">
<label for="materials">Materials: </label>
<input id="materials" />
</div>
</div><!-- End demo -->
</body>
</html>它工作得很完美。太好了,事实上,我想我会保留这段代码,不是它应该如何工作,但是……
发布于 2012-07-19 17:13:08
你的PHP完全错了:
while ($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.
}应该是:
$cname = array();
while ($rs = mysqli_fetch_array($rsd)) {
$cname[]['label'] = $rs['name']; // I know this all returns correctly
break;
}
echo json_encode($cname); // First time I have ever used json, error might be here.label是jqueryautocomplete (我相信)使用的数组行中的默认标签字段。此外,返回的必须是一个数组数组,每个数组行代表一个匹配项。
您可以通过添加一个值字段来使其变得更复杂,以便通过执行以下操作使文本框实际等于:
$cname = array();
while ($rs = mysqli_fetch_array($rsd)) {
$cname[]['label'] = $rs['name']; // I know this all returns correctly
$cname[]['value'] = $rs['id'];
break;
}
echo json_encode($cname); // First time I have ever used json, error might be here.当然,我不认为你真的想要我放进去的break;,因为:
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.
}向我表示您实际上从结果中返回了一行。如果不是,实际上是返回所有结果,则取出break;。
发布于 2012-07-19 16:46:24
在PHP部分,也许可以尝试这样的东西:
$res = array(); //create a new array
while ($rs = mysqli_fetch_array($rsd)) {
$res[] = (string)$rs['name']; //add results to array, casted as string
}
header('Content-type: application/json'); //add JSON headers (might work w/o)
echo json_encode($res); //output array as JSON你应该像['name1', 'name2', 'name3']一样把所有的结果放在一个数组中,这是...that的方式
发布于 2012-07-19 16:48:53
试试这段代码,它对我很有效。
$().ready(function() {
$("#materials").autocomplete("autocomplete.php", {
width: 260,
matchContains: true,
autoFill:true,
selectFirst: false
});
});https://stackoverflow.com/questions/11557003
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