php - 在新的虚拟主机上出现此错误 undefined offset : 1 and 2

Question

我最近将我的脚本移到了一个新的虚拟主机中，现在我遇到了这个错误

注意: undefined offset :1($id2 的行)注意: undefined offset :2($id3 的行)

这是我的 PHP 代码

<?php
include '../connect_database.php'; 

$sql = "SELECT score FROM dailyscore Where Id IN (1,2,3)";
date_default_timezone_set('America/New_York');
$result = $connect->query($sql);
while ($row = mysqli_fetch_assoc($result)){
$rows[] = $row; 
$id1= $rows[0]['score'];
$id2= $rows[1]['score'];
$id3= $rows[2]['score'];

}

$list['scores'] = array('data' => $id1, 'data1' => $id2, 'data2' => $id3);

$myJSON = json_encode($list);

echo $myJSON;
print_r($rows);
?>

知道为什么吗？

Answer 1

我认为这是设计缺陷的表现。看起来您正在尝试查找 3 位玩家的分数。您正在遍历 3 行，但试图在每次迭代中访问所有 3 名玩家的数据。相反，您应该在各自的迭代中访问每个玩家的数据，并建立一个玩家数据列表。

要直接回答您的问题，在迭代 1 中您尝试访问元素 0、1 和 2，但 $rows 仅填充了 0。

+-----------+-------------------------+--------------------------+
| Iteration | You're trying to access | You only have access to  |
+-----------+-------------------------+--------------------------+
|         1 |                   0,1,2 |                        0 |
|         2 |                   0,1,2 |                      0,1 |
|         3 |                   0,1,2 |                    0,1,2 |
+-----------+-------------------------+--------------------------+

例子

<?php
// Turn on error reporting
ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);

// Show MySQL errors as PHP exceptions
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

include '../connect_database.php'; 

// Build up this list inside the loop. 
$scores = [];

$sql = "SELECT score FROM dailyscore Where Id IN (1,2,3)";
date_default_timezone_set('America/New_York');
$result = $connect->query($sql);
while ($row = mysqli_fetch_assoc($result)){
    // You only have access to a single $row here. 
    // Build up an array of the data you want instead of tring to access "other rows"
    $scores[] = $row['score'];
}

// Now, you can use $scores to build $list... or build $list inside the loop. 
?>

编辑

Do you mind to show me example how to assign the array results to something like this? $list['scores'] = array('data' => $id1, 'data1' => $id2, 'data2' => $id3);

<?php
// Turn on error reporting
ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);

// Show MySQL errors as PHP exceptions
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

include '../connect_database.php'; 

// Build up this list inside the loop. 
$scores = [];
$counter = 0;

$sql = "SELECT score FROM dailyscore Where Id IN (1,2,3)";
date_default_timezone_set('America/New_York');
$result = $connect->query($sql);
while ($row = mysqli_fetch_assoc($result)){

    $keyName = 'data';

    if ($counter > 0) {
        $keyName = 'data' . $counter;
    }

    $scores[$keyName] = $row['score'];

    $counter++;
}

$list['scores'] = $scores;

echo json_encode($list);