Php 将MYSQL SELECT转换为图表数据的JSON

我在一个表上执行select，在表中检查列中与中的数据相对应的最高值，这很好，但是我需要将这些值转换为JSON，以便在图表中使用这些值

<?php


    $result = $conn->query("SELECT MAX(score_1) `top_score_1` FROM members WHERE dashboard_id = $user_dash_id");      
    if (!$result) die($conn->error);
    while($row=mysqli_fetch_array($result, MYSQLI_ASSOC)) 
    {
        $top_score_1 =  $row['top_score_1'];
    }

    $result = $conn->query("SELECT MAX(score_2) `top_score_2` FROM members WHERE dashboard_id = $user_dash_id");      
    if (!$result) die($conn->error);
    while($row=mysqli_fetch_array($result, MYSQLI_ASSOC)) 
    {
        $top_score_2 =  $row['top_score_2'];
    }

    $result = $conn->query("SELECT MAX(score_3) `top_score_3` FROM members WHERE dashboard_id = $user_dash_id");      
    if (!$result) die($conn->error);
    while($row=mysqli_fetch_array($result, MYSQLI_ASSOC)) 
    {
        $top_score_3 =  $row['top_score_3'];
    }

    $result = $conn->query("SELECT MAX(score_4) `top_score_4` FROM members WHERE dashboard_id = $user_dash_id");      
    if (!$result) die($conn->error);
    while($row=mysqli_fetch_array($result, MYSQLI_ASSOC)) 
    {
        $top_score_4 =  $row['top_score_4'];
    }

    $result = $conn->query("SELECT MAX(score_5) `top_score_5` FROM members WHERE dashboard_id = $user_dash_id");      
    if (!$result) die($conn->error);
    while($row=mysqli_fetch_array($result, MYSQLI_ASSOC)) 
    {
        $top_score_5 =  $row['top_score_5'];
    }

    $result = $conn->query("SELECT MAX(score_6) `top_score_6` FROM members WHERE dashboard_id = $user_dash_id");      
    if (!$result) die($conn->error);
    while($row=mysqli_fetch_array($result, MYSQLI_ASSOC)) 
    {
        $top_score_6 =  $row['top_score_6'];
    }

    $result = $conn->query("SELECT MAX(score_7) `top_score_7` FROM members WHERE dashboard_id = $user_dash_id");      
    if (!$result) die($conn->error);
    while($row=mysqli_fetch_array($result, MYSQLI_ASSOC)) 
    {
        $top_score_7 =  $row['top_score_7'];
    }

    $result = $conn->query("SELECT MAX(score_8) `top_score_8` FROM members WHERE dashboard_id = $user_dash_id");      
    if (!$result) die($conn->error);
    while($row=mysqli_fetch_array($result, MYSQLI_ASSOC)) 
    {
        $top_score_8 =  $row['top_score_8'];
    }                                                

    $total_five_stars = $top_score_1 + 
                        $top_score_2 + 
                        $top_score_3 + 
                        $top_score_4 + 
                        $top_score_5 + 
                        $top_score_6 + 
                        $top_score_7 + 
                        $top_score_8;
    ?>

---

如果我理解正确，您希望从php脚本获得json格式的响应。在这种情况下，功能和将帮助您完成此任务

$data = [
    $top_score_1, $top_score_2, $top_score_3, $top_score_4,
    $top_score_5, $top_score_6, $top_score_7, $top_score_8
];

header('Content-Type: application/json');
echo json_encode($data);

$data = [
    $top_score_1, $top_score_2, $top_score_3, $top_score_4,
    $top_score_5, $top_score_6, $top_score_7, $top_score_8
];

header('Content-Type: application/json');
echo json_encode($data);