如何在数据库中插入2个变量的商，然后将其作为百分比回显

发表时间：2022-07-31 01:43:26 阅读：189

$score = 50
$total = 500

mysqli_query($conn,"UPDATE student_quiz SET `grade` = ".$score." out of ".$total."')

因此，目前这将在数据库中存储"500分之50"，但我想它做的是存储$score和$total的商，因此它应该是"0.1"，而不是"500分50".

当我显示它时，我如何将其作为百分比，这样当我回显$grade时，它将变为"10%"，而不是"0.1".目前，这是我获取和显示分数的方式.

$query1 = mysqli_query($conn,"select * from student_class_quiz where class_quiz_id = '$id' and student_id = '$session_id'")or die(mysqli_error());

    $row1 = mysqli_fetch_array($query1);
    $grade = $row1['grade'];
   <?php echo $grade; ?>

🎖️ 优质答案

用`$score/$total计算商.

$score = 50;
$total = 500;
$quotient = $score/$total;
$stmt = mysqli_prepare($conn,"UPDATE student_quiz SET `grade` = ?");
$stmt->bind_param("s", $quotient);
$stmt->execute();

要将其显示为百分比，请乘以100.

<?php
$query1 = mysqli_prepare($conn,"select * from student_class_quiz where class_quiz_id = ? and student_id = ?") or die(mysqli_error($conn));
$query1->bind_param("ss", $id, $session_id);
$query1->execute();
$result = $query1->get_result();
$row1 = $result->fetch_assoc($query1);
$grade = $row1['grade'];
echo "Grade percentage is " . round($grade*100) . "%";