php - 如何从传递给 php 查询请求的 jquery 获取 VAR

Question

<script src="http://code.jquery.com/jquery-1.5.min.js"></script>
<script>
$(document).ready(function() {
    var $article = null;

    $('#category').change(function() {
        var $categoryName = $('#category').val();

        if ($article == null) {
            $article = $('<h4>Select a business you wish to view.</h4><select id="business" name="business"   class="business"><option value="0">Select A Business To View Listing</option></select>').appendTo('.query');
        }
        $("#business").load("php.php");
    });
});
</script>

这就是我当前正在做的事情，我正在加载 php.php 脚本，而是想将 $categoryName 的值传递给查询中的 WHERE 子句，如下所示:

<?php
    $con = mysqli_connect(,,,,);
    // Check connection
    if (mysqli_connect_errno())
       {
       echo "<option>Failed to connect to MySQLi</option>" ;
       }
    $result = mysqli_query($con,"SELECT Bname, Category FROM Business WHERE Category='$categoryName'");
     while($row = mysqli_fetch_array($result)) {
      echo "<option value='".$row['Bname']."'>".$row['Bname']."</option>";
      }
      // Free result set
     mysqli_free_result($result);
     mysqli_close($con);
?>

这应该工作的方式是，第一个选择框由我的服务器上的 php 填充，显示类别列表。用户从该框中选择一个类别，然后在更改时，创建第二个选择框，将其添加到表单中，并将进行查询，以填充我的数据库中列出的共享类别的所有企业名称(即第一个框中的选定索引) )你能帮我改变它，使其按照我需要的方式工作吗？

更新:这是更新后的代码，现在第二个选择框永远不会加载。

 <script src="http://code.jquery.com/jquery-1.5.min.js"></script>
 <script>
 $(document).ready(function () {
 var $article = null;

 $('#category').change(function () {
 var categoryName = $('#category').val();

 if ($article == null) {
 $article = $('<h4>Select a business you wish to view.</h4><select id="business" name="business" class="business"><option value="0">Select A Business To View Listing</option></select>').appendTo  ('.query');

  $("#business").load( "php.php",
     data:{myVar:$categoryName}
  );
  }
 });
 });
</script>

这是 php.php

 <?php
    $con = mysqli_connect(,,,,);
    // Check connection
    if (mysqli_connect_errno())
       {
       echo "<option>Failed to connect to MySQLi</option>" ;
       }
    $myVar = $_GET["myVar"];
    $result = mysqli_query($con,"SELECT Bname, Category FROM Business WHERE Category='$myVar'");
     while($row = mysqli_fetch_array($result)) {
      echo "<option value='".$row['Bname']."'>".$row['Bname']."</option>";
      }
      // Free result set
     mysqli_free_result($result);
     mysqli_close($con);
 ?>

Answer 1

如下所示，您可以使用一个参数 myVar 创建一个 data 对象JS:

$('#category').change(function() {
    var $categoryName = $('#category').val();

    if ($article == null) {
        $article = $('<h4>Select a business you wish to view.</h4><select id="business" name="business"   class="business"><option value="0">Select A Business To View Listing</option></select>').appendTo('.query');
    }
    $("#business").load("php.php",
         data:{myVar:$categoryName}
    );
});

然后像这样在PHP中获取变量PHP:

$con = mysqli_connect(,,,,);
$myVar = $_GET["myVar"];
// Check connection
if (mysqli_connect_errno())
   {
   echo "<option>Failed to connect to MySQLi</option>" ;
   }