Question

我想要做的是通过jQueryUI的功能使用自动完成分类结果。经过一些谷歌搜索等我发现它有一个内置函数（http://jqueryui.com/demos/autocomplete/#categories），但该示例仅适用于本地数据源（javascript中的数组）。我正在处理远程数据源。我的代码是

<script>
$( function() {
 $.widget("custom.catcomplete", $.ui.autocomplete, {
_renderMenu: function(ul, items) {
    var self = this,
        currentCategory = "";
    $.each(items, function(index, item) {
        if (item.category != currentCategory) {
            ul.append("<li class='ui-autocomplete-category'>" + item.category + "</li>");
            currentCategory = item.category;
        }
        self._renderItem(ul, item);
    });
}
});
$( "#search" ).catcomplete({
    delay:0,
    source: "search.php",
    select: function(event, ui){
    alert(ui.item.label);
}
});
} );
 </script>
 </head>
 <body>
<label for="search">Search: </label>
<input id="search">
</body>

这里是 search.php

<?php
 $conn = mysqli_connect("localhost","root","","test") or die(mysqli_error());
 $searchTerm = $_GET['term'];
 $sql = "select * from country_ref_table where country_code LIKE '%".$searchTerm."%' ORDER BY country_code ASC";
 $result = mysqli_query($conn,$sql) or die(mysqli_error($conn));
 $data = [];
 while($row = mysqli_fetch_array($result))
  {
   $data[] = $row['country_code'];
   $data[] = $row['country'];
  }
 echo json_encode($data);
  ?>

Answer 1

 $searchTerm = $_GET['term'];
 $sql = "select * from country_ref_table";

将其添加到$ sql

中

 $sql = "select * from country_ref_table where `term` LIKE '%".$searchTerm ."%'";

然后使用这样的数据：

$data=array();
while($row = mysqli_fetch_array($result))
  {
   $data[]['id'] = $row['country_code'];
   $data[]['category'] = $row['country'];
  }
 echo json_encode($data);

Answer 2

而不是：

 $searchTerm = $_GET['term'];
 $sql = "select * from country_ref_table";

您需要在查询中加入$searchTerm，有点像：

$searchTerm = $_GET['term'];

$sql = "select * from country_ref_table WHERE `columnTerm` LIKE '%".$searchTerm ."%' ";

OR

$sql = "select * from country_ref_table WHERE `columnTerm` = '$searchTerm' ";

在php mysql中使用类别自动完成功能无效

2 个答案: