Question

我正在尝试将动态下拉框作为搜索工具，以帮助缩小mysql服务器的显示数据。我是一个体面的PHP程序员，但需要javascript和ajax的帮助。

该网站目前包含3个页面：index_test.php，dropdown.php和dropdown2.php。

在index_test.php上有4个下拉菜单需要填充信息。第一个是在页面加载时使用php从mysql表填充状态名称。第二个框使用引用php代码的.change（）填充，并从mysql表中显示处于选定状态的学校。然后，第三个框应该从第二个框中获取所选值，并将所选学校的类名显示给用户，该步骤是代码中断的位置。 php通过提交表单进行测试时工作，但我希望能够在没有页面刷新的情况下填充最后2个框。 mysql表的格式是：表学校:( school_id，学校，州）表类：（class_id，school_id，class_abrv，class_number）

感谢您的帮助

index_test.php的代码：

<?php include_once("connect.php"); ?>
<html>
<head>
<title>ajax</title>
<script src="jquery.js"></script>

<script type="text/javascript">
$(document).ready(function(){
$("#state").change(function(){
var state = $("#state").val();
$.ajax({
type:"post", 
url:"dropdown.php", 
data:"state="+state, 
success: function(data) {
$("#school").html(data);
}
});
});

$("#school").change(function(){
var state = $("#school").val();
$.ajax({
type:"post", 
url:"dropdown2.php", 
data:"school="+school, 
success: function(data) {
$("#classname").html(data);
}
});
});

});

</script>


</head>
<body>

<h1>Get Notes:</h1>
<br/>
<form action="dropdown2.php" method="post">

State: <select id="state" name="state">
<option>--Select State--</option>
<?php   
$sql = "SELECT states FROM states";
$result = mysql_query($sql);

while ($output = mysql_fetch_array($result)) {
$state_name = $output['states'];
echo "<option value=\"$state_name\">$state_name</option>";
}       
?>
</select>
<br/>

School: <select id="school" name="school">
<option>--Select School--</option>
</select>
<br/>

Class Name: <select id="classname" name="classname">
<option>--Select Class Name--</option>
</select>
<br/>

Class Number: <select id="classnumber" name="classnumber">
<option>Select Class Name</option>
</select>
<br/>

<input type="submit" value="Search" />
</form> 


</body>
</html>

Dropdown.php：

<?php
include_once("connect.php");

$state=$_POST["state"];
$result = mysql_query("select schools FROM schools where states='$state' ");
while($school = mysql_fetch_array($result)){
echo"<option value=".$school['schools'].">".$school['schools']."</option>";
}
?>

Dropdown2.php

<?php
include_once("connect.php");

$school=$_POST['school'];
$result = mysql_query("SELECT school_id FROM schools WHERE schools='$school' ");
$school_id = mysql_fetch_array($result);

$id = $school_id['school_id'];
$classname = mysql_query("SELECT DISTINCT class_abrv FROM classes WHERE school_id='$id'           ORDER BY class_abrv asc");

while($class = mysql_fetch_array($classname)){
echo"<option value=".$class['class_abrv'].">".$class['class_abrv']."</option>";
} 
?>

Answer 1

在第二个ajax函数中，您已将学校下拉框值分配给状态变量，但是您将变量school传递给ajax帖子。所以没有学校变量就是你得错的原因。

$("#school").change(function(){
 var *state* = $("#school").val();
 //above variable should be school.
 $.ajax({
  type:"post", 
  url:"dropdown2.php", 
  data:"school="+*school*, 
  success: function(data) {
   $("#classname").html(data);
  }
 });
});

动态依赖下拉菜单与ajax php mysql

1 个答案: