我在填充下拉列表时遇到问题。
我有一个用户选择分支的下拉列表,以及显示相关选项的第二个下拉列表。
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>';
}
答案 0 :(得分:1)
以下是您的问题http://www.w3schools.com/php/php_ajax_database.asp
的确切答案如果将此示例集成到您的代码中,则应该是:
<强> HTML 强>
php 1.php -xmycommand -d"my option"
<强> JS 强>
var_dump(getopt('x:d:'));
您应该将文件“test.php”的名称替换为此行 xmlhttp.open(“GET”,“test.php?branch =”+ branch,true); < / p>