我正在尝试根据他们的班级列出学生,当我选择班级名称并刷新表格时,它不会列出学生并给出致命错误
致命:在布尔值中调用成员函数fetch_assoc() C:\ ... XAMPP
这是我编写的功能,我想根据他们的班级选择学生:
function std_list_header(){
$Connection=new mysqli("localhost","root","","dar_ul_hidaya");
$query =" SELECT * FROM class";
$res = mysqli_query($Connection, $query);
if($res){
$rs=mysqli_fetch_assoc($res);
echo '<tr><td colspan="2" style="border-right:none;"><b>Class:'.$rs["jamat"].'</b></td>
<td colspan="3" >
<form action="student.php" method="POST">
<select name="jamat">
<option value="-1">Select Class</option>';
$jamat="";
if (isset($_POST["jamat"])) {
if($_POST["jamat"]>=0){
$jamat=$_POST["jamat"];
}
}
$query="SELECT jamat FROM class";
//echo $query;
$result=$Connection->query($query);
while($rs=$result->fetch_assoc()){
echo"<option value=".$rs["id"]." ".($jamat==$rs["id"]?"selected":"").">".$rs["jamat"]."</option>";
}
echo '</select>
<input type="submit" value="Refresh">
</form>
</td>
</tr>';}
}
这是我显示学生名单的主页:
<table align="center" border="1px" width="600px">
<?php
std_list_header();
?>
<tr>
<th width="30px">کلاس</th>
<th width="30px">ولدیت</th>
<th width="30px">نام</th>
<th width="30px">نمبر شمار</th>
</tr>
<?php
$query="SELECT student.id ,student.name ,student.fname,student.class FROM
student
inner join enroll on student.id=enroll.student_id
inner join class on enroll.class_id=class.id";
if (isset($_POST["jamat"])) {
if($_POST["jamat"]>=0){
$query=$query." where class.id=".$_POST["jamat"];
};
}
$resultset=$Connection->query($query);
$count=0;
while($r=$resultset->fetch_assoc()){
$count++;
echo "
<tr>
<td>".$r["class"]."</td>
<td>".$r["fname"]."</td>
<td>".$r["name"]."</td>
<td>".$count."</td>
</tr>
";
}
?>
</table>
</tr>
</table>