【发布时间】:2017-04-29 10:46:52
【问题描述】:
代码:
<script>
$(document).ready(function(){
$(".field").change(function(){
field = $(".field").val();
$.ajax({
type:"POST",
data:{"field":field},
url:"potential-courses.php",
success:function(data){
$(".course").val(data);
}
});
});
});
</script>
潜在课程.php
<?php
include("conn.php");
$field = $_POST['field'];
$sql = "select * from course_master where field = '$field' order by course_full_name";
$result = mysqli_query($link,$sql);
while ($row = mysqli_fetch_array($result))
{
echo "<option value=".$row['course_short_name'].">".$row['course_full_name']."</option>";
}
?>
html代码:
<select name='field' class='field' id="field">
<option value="">Select Field</option>
<option value='engineering'>Engineering</option>
<option value='law'>LAW</option>
<option value='medical'>Medical</option>
<option value='management'>Management</option>
<option value='pharmacy'>Pharmacy</option>
<option value='hotel management'>Hotel Management</option>
<option value='mass communication'>Mass Communication</option>
<option value='agriculture'>Agriculture</option>
<option value='architecture'>Architecture</option>
<option value='education'>Education</option>
<option value='paramedical'>Paramedical</option>
<option value='design'>Design</option>
<option value='commerce'>Commerce</option>
<option value='film/tV/media'>Film /TV/ Media</option>
</select>
<select name="course" class="course">
<option value="">Select Courses</option>
</select>
在这段代码中,我有两个下拉列表,即
<select name='field' class='field' id="field">
另一个是
<select name="course" class="course">
当我从“name=field”更改值时,它在“name=course”中不显示任何内容。我在哪里做错了请帮助我。
谢谢
【问题讨论】:
-
控制台返回什么ajax?尝试在 ajax 成功时提醒数据。