【发布时间】:2016-05-23 11:29:55
【问题描述】:
我正在尝试使用 jquery 自动完成功能在 input 字段中显示用户列表。我在向字段显示名称和选择更新值时遇到问题。
我的 PHP 代码:
include '../_db.php';
// get what user typed in autocomplete input
$name = trim($_GET['name']);
$param = "%{$name}%";
$query = $conn->prepare('SELECT emp_number, emp_firstname, emp_lastname FROM `hs_hr_employee` WHERE emp_firstname LIKE ? OR emp_lastname LIKE ? ');
$query->bind_param('ss', $param,$param);
$query->execute();
$query->bind_result($emp_number,$emp_firstname,$emp_lastname);
$a_json = array();
$a_json_row = array();
while( $query->fetch() )
{
$a_json_row["emp_number"] = $emp_number;
$a_json_row["fname"] = $emp_firstname;
$a_json_row["lname"] = $emp_lastname;
array_push($a_json, $a_json_row);
}
$json = json_encode($a_json);
print $json;
$query->close();
我的 JS 代码
$(function()
{
$( "#search-emp" ).autocomplete(
{
source: function (request, response)
{
var form_data = {
ajax : '1',
name : $("#search-emp").val(),
actioncall : 'search-emp'
};
$.ajax({
type: "POST",
url: "_ajax.php",
data: form_data,
success: function(response)
{
$.each( response, function( key, value )
{
//alert( key + ": " + value );
console.log('element at index ' + key + ' is ' + JSON.parse(value));
});
//console.log(response);
},
dataType: 'json'
});
}
}, {minLength: 3 });
});
得到响应
[{"emp_number":1,"fname":"Arslan","lname":"Hassan"},{"emp_number":2,"fname":"Muneeb","lname":"Janjua" },{"emp_number":3,"fname":"hr","lname":"user"}]
我的 HTML 代码
<form class="form-inline">
<div class="form-group">
<label for="exampleInputName2">Employee Name: </label>
<input id="search-emp" type="text" class="form-control" placeholder="*">
</div>
<div class="form-group">
<label for="exampleInputEmail2">Date Range: </label>
<input type="text" class="form-control" id="dp-from" placeholder="From">
<input type="text" class="form-control" id="dp-to" placeholder="To">
</div>
<button type="submit" class="btn btn-primary">Genrate Report</button>
</form>
【问题讨论】:
标签: php jquery autocomplete