这是大学的小型次要项目。它根据票数对学生进行排名。它确实在分页中对学生进行排名,但是当单击学生资料时,排名不会显示。
好的,下面的代码可以工作,但是代码的最后部分不显示输出中的排名。我如何在两个代码中都使用$ num?
<a href="/">HOME</a>
<br><br>
<form>
<input type="text" name="id" autocomplete="off" placeholder="enter student id">
</form>
<br><br>
<?php include 'conn.php'; ?>
<?php
$sql = "SELECT * FROM table";
$result = mysqli_query($conn, $sql);
$result_per_page = 5;
$no_of_result = mysqli_num_rows($result);
$no_of_page = ceil($no_of_result/$result_per_page);
if(!isset($_GET['page'])){
$page = 1;
}else{
$page = $_GET['page'];
}
$page_first = ($page-1)*$result_per_page;
$sql = 'SELECT * FROM table ORDER BY votes DESC LIMIT ' . $page_first . ',' . $result_per_page;
$result = mysqli_query($conn, $sql);
$num = $page * $result_per_page -4;
while($row = mysqli_fetch_assoc($result)) {
echo '<div class="x">';
echo '#';
echo $num++;
echo ' ';
echo '<a href="?id='.$row['id'].'">';
echo $row['name'];
echo '</a>';
echo '</div>';
}
echo '<br>';
for($page=1;$page<=$no_of_page;$page++){
echo '<a href="/?page='.$page.'">'.$page.'</a>';
echo ' ';
}
echo '<br>';
?>
<?php
$name = $_GET['id'];
$sql = "SELECT * FROM table WHERE id=$name";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
echo '<br>';
echo 'RANK = ';
echo '?';//display rank here?
echo '<br>';
echo 'NAME = ';
echo $row['name'];
echo '<br>';
echo 'VOTES = ';
echo $row['votes'];
echo '<br>';
}
} else {
//echo "0 results";
}
?>
以上代码的输出。我想在问号处输出排名。
current output
我想要这样的东西
desired output