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我对 AJAX / jQuery 还很陌生,我只是在探索这是我的问题我有一个文本框,我想要做的是如果我单击搜索按钮,搜索结果将加载(#sort-results)我不知道这有什么问题..我希望有人能让我理解更多。谢谢。

索引.php

<input type="text" name="item_type" class="item_type"></td>
        <select class="location">
          <option value="bldg1">bldg1</option>
          <option value="bldg2">bldg2</option>
        </select>
        <button class="btn_search">Search</button>

      <div id="sort-results"></div>

脚本.js

$(".btn_search").click(function(){
  type = $(".item_type").val();
  location = $(".location").val();
    if(type.length > 0){
    $.ajax({
      type: "GET",
      url: "sort.php",
      data: "type="+type +"location="+location,
      cache:false,
      success : function(){
        $("#sort-results").html();
      }

    });
   }

});

排序.php

require 'global/db.php';
require_once 'config.php';
$type = $_GET['type'];
$location = $_GET['location'];

$db = new Database(DBHOST, DBUSER, DBPASS, DB);
$selection = $db->select("SELECT * FROM os_inventory WHERE os_type like '$type' AND os_location like '$location'");
echo "<table class=bordered>
        <tr><th>Item</th>
        <th>Item Code</th>
        <th>Item Type</th>
        <th>Serial No.</th>
        <th>Date Added</th>
        </tr>";
foreach ($selection as $key) {
    echo "<tr>";
    echo "<td>" . $key['os_id'] . "</td>";
    echo "<td>" . $key['os_type'] . "</td>";
    echo "<td>" . $key['os_code'] . "</td>";
    echo "<td>" . $key['os_serial'] . "</td>";
    echo "<td>" . $key['os_date'] . "</td>";
    echo "</tr>";
}
echo "</table>";
4

2 回答 2

1

您是否忘记实际传递内容?

success : function(html){
             $("#sort-results").html(html);
          }
于 2013-08-18T06:47:58.657 回答
0

也许我错了,但我没有看到 jQuery 选择器使用句点。将您的选择器更改为:

$("#btn_search").click(function(){
      type = $("#item_type").val();
      location = $("#location").val();

然后按照@David的帖子实际从服务器接收代码以放入您的 DIV

于 2013-08-19T14:51:29.323 回答