我是php的新手，我正在尝试创建一个学校门户，现在我正在尝试创建一个页面，管理员可以在该页面上课程。我只需要知道如何做，就是允许管理员从下拉菜单中选择一位老师（名称或用户名直接来自数据库），然后在Teacher_id的“班级”表中输入相应的ID。
$query = "SELECT Username FROM tbluser WHERE Role='Teacher'";
$result2= mysqli_query($conn, $query);
<body>
<form action="createclass.php" method="POST">
<div>
<label>Class name</label>
<input type="text" name="classname" class="form-control" placeholder="Enter new username">
<br/>
<label>Teacher</label>
<select name ="teacher">
<option selected = "true" disabled="disabled"> Select one from below...</option>
<?php
while ($rows=mysqli_fetch_assoc($result2))
{
$teachername= $rows['Username'];
echo"<option value ='$teachername'>$teachername</option>";
}
?>
</select>
if(isset($_POST["btnAddclass"])){
$classname = $_POST["classname"];
$teacher = $_POST["teacher"];
$subject = $_POST["subject"];
$section = $_POST["section"];
$yeargroup = $_POST["yeargroup"];
$result1 = mysqli_query($conn, "SELECT ID FROM teacher INNER JOIN tbluser ON tbluser.ID=teacher.tbluser_ID WHERE Username = '$teacher'"); // using mysqli_query instead
$sql = "INSERT INTO class (classname, teacher_id, yeargroup_id, subject_id, section_id) VALUES ('$classname', '$result1', '$subject','$section','$yeargroup')";
if(!mysqli_query($conn,$sql))
{
echo "Error";
}
else
{
echo "New class has been added.";
}
}
也从tbluser中选择ID，该ID是教师ID：
$query = "SELECT ID, Username FROM tbluser WHERE Role='Teacher'";
$result2= mysqli_query($conn, $query);
<form action="createclass.php" method="POST">
<div>
<label>Class name</label>
<input type="text" name="classname" class="form-control" placeholder="Enter new username">
<label>Teacher</label>
<select name="teacher">
<option selected="true" disabled="disabled"> Select one from below...</option>
<?php
while ($rows=mysqli_fetch_assoc($result2)) {?>
<option value="<?php echo $rows['ID']?>"><?php echo $rows['Username']?></option>
<?php
}?>
</select>
提交时：
$classname = $_POST["classname"];
$teacherID = $_POST["teacher"];
$subject = $_POST["subject"];
$section = $_POST["section"];
$yeargroup = $_POST["yeargroup"];
$sql = "INSERT INTO class (classname, teacher_id, yeargroup_id, subject_id, section_id) VALUES ('$classname', '$teacherID', '$subject','$section','$yeargroup')";
并且当您使用来自用户输入的不安全数据时，请考虑移至准备的语句以避免SQL注入。