我有一个html表单,PHP scrip和jquery.我需要一个Ajax代码来从我的PHP脚本中自动提出建议.以下是代码…
Form.html
<html>
<head>
<script src="jquery1.6.4.min.js" type="text/javascript"></script>
<script src="jquery.jSuggest.js" type="text/javascript"></script>
<link href="jSuggest.css" rel="stylesheet" type="text/css" />
</head>
<body>
<form id="form1" name="form1" method="post" action="#">
<input type="text" name="TagsInputField" id="TagsInputField"/>
</form>
</body>
</html>
test.PHP的
<?PHP
include("bc/script/core/dbcon.PHP");
$input = $_POST['TagsInputField'];
$data = array();
// query your DataBase here looking for a match to $input
$query = MysqL_query("SELECT * FROM user WHERE username LIKE '%$input%'");
while ($row = MysqL_fetch_assoc($query)) {
$json = array();
$json['value'] = $row['id'];
$json['name'] = $row['username'];
$data[] = $json;
}
header("Content-type: application/json");
echo json_encode($data);
?>
jquery.jSuggest.js
$(function() {
var dataSource = {
items: [
{
value: "21",
name: "Mick Jagger"},
{
value: "43",
name: "Johnny Storm"},
{
value: "46",
name: "Richard Hatch"},
{
value: "54",
name: "Kelly Slater"},
{
value: "79",
name: "Michael Jordan"}
]
};
$('#TagsInputField').jSuggest({
source: dataSource.items,
selectedItemProp: "name",
seekVal: "name",
selectionAdded: function(elem, data) {
console.log(data.name);
},
selectionRemoved: function(elem, data) {
console.log(data.name);
elem.remove();
}
});
});
请注意,指针“源”是指对象“ dataSource.items”以读取建议.
谁能帮我写一个Ajax代码来读取建议的PHP文件,该文件返回一个json.
解决方法:
type: "POST"
在规则中.
您的jSuggest规则中还有其他一些主要错误.您应该阅读文档：http://scottreeddesign.com/project/jsuggest
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