Question

我有以下代码调用jquery自动完成小部件：

$(function() {
        $( "#vendor_name" ).autocomplete({
        source: 'vendor_names.php',
        minLength: 3
})
});

我的vendor_names.php文件类似于：

<?php
include("include/db_connect.php");

$query = "select VendorName from Vendor where VendorCancelDate is NULL order by 
VendorName";
$result = mssql_query($query);
while ( $record = mssql_fetch_array($result) ){
        $vendors[] = array('label' => $record['VendorName']);
}
echo json_encode($vendors);

?>

但是当我输入任何内容时，它总是返回查询中的所有内容。有什么想法吗？

Answer 1

将输入值传递给脚本：

$(function() {
        var val = $( "#vendor_name" ).val();
        $( "#vendor_name" ).autocomplete({
        source: 'vendor_names.php?token=' + val,
        minLength: 3
   });
});

过滤脚本返回的结果。您将希望以不连接sql以避免sql注入的方式执行此操作，但仅作为示例：

<?php
include("include/db_connect.php");
$val = $_GET["token"];
$query = "select VendorName from Vendor where VendorCancelDate is NULL and VendorName like '%". $val . "%' order by VendorName";
$result = mssql_query($query);
while ( $record = mssql_fetch_array($result) ){
        $vendors[] = array('label' => $record['VendorName']);
}
echo json_encode($vendors);

?>

Answer 2

据我所知，您需要将用户的输入传递给vendor_names.php?input=whatever ..
并且您的查询必须是
select VendorName from Vendor where VendorCancelDate is NULL and VendorName=%whatever%' order by VendorName

jquery autocomplete返回所有标签

2 个答案: