【发布时间】:2016-02-24 05:29:57
【问题描述】:
我在填充下拉列表时遇到问题。
我有一个下拉菜单供用户选择分支,还有一个下拉菜单显示相关选项。
HTML
<select id="first-choice" onchange="leaveChange()">
<option selected value="base">Please Select</option>
<option value="CSE">CSE</option>
<option value="ECE">ECE</option>
<option value="EEE">EEE</option>
<option value="MECH">MECH</option>
</select>
<br>
<select id="second-choice">
<option>Please choose from above</option>
</select>
JS
function leaveChange(){
//what to insert here;
}
PHP
$branch=$_GET['branch'];
$username = "jaggu";
$password = "8374";
$hostname = "localhost";
$dbhandle = mysqli_connect($hostname, $username, $password) or die("Unable to connect to MySQL");
$selected = mysqli_select_db($dbhandle,"test") or die("Could not select examples");
$query = "SELECT * FROM `register` WHERE classid LIKE '%".$branch."%'";
$result = mysqli_query($dbhandle,$query);
$row = mysqli_fetch_object($result);
$query1="SELECT * FROM `examdup` WHERE `classid` LIKE '%".$branch."%'";
$result1= mysqli_query($dbhandle,$query1);
while ($row1= mysqli_fetch_object($result1)) {
echo '<option value="'.$row1->month_year.'">'. $row1->title.'</option>';
}
【问题讨论】:
-
看看this。你也可以谷歌
Ajax populate dropdown -
无法弄清楚先生,我是这类事情的初学者,这就是为什么我发布了两个代码我下来
标签: javascript php drop-down-menu html-select