打乱后的结果未按打乱后的顺序发送到数据库

Question

我有一个页面正在重新排列用户订单。我想将该订单的结果发送到我的数据库中，以便我以后可以将该订单用于某些事情。我认为结果是按照它们在洗牌时出现的顺序发送的，但事实并非如此，我不确定它们发送的确切顺序。

$query = mysqli_query($con, "SELECT * FROM users WHERE `group` IN (3, 4 ,5)");

echo 'Users to be given draft order: <br>';
$array = array();
while ($row = mysqli_fetch_assoc($query)) {
    $array[] = $row;
    echo $row['firstname'] . ' ' . $row['lastname'] . '<br>';
}
?>
<form method="

POST" name="form">
    <input type="submit" value="Create Draft Order" name="shuffle">
     <input type="hidden" name="shuffleButton" value="1">
</form>

Shuffled results: <br>
<div class="main-bag">     
<div class="shuffle_results" id="results"></div> 
 <form method="post">
    <input type="submit" value="Finalize Draft Order" name="insert">
<?php
$count = 0;
foreach ($array as $result) :
$count++;
$shuffle_count = $count;
    $shuffle_firstname = htmlentities($result['firstname']);
    $shuffle_lastname = htmlentities($result['lastname']);
    $shuffle_id = htmlentities($result['id']);
    $shuffle_username = htmlentities($result['username']);
    $shuffle_email = htmlentities($result['email']);
?>
    <input type="hidden" name="count[]" value="<?php echo $shuffle_count; ?>">
        <input type="hidden" name="firstname[]" value="<?php echo $shuffle_firstname; ?>">
        <input type="hidden" name="lastname[]" value="<?php echo $shuffle_lastname; ?>">
        <input type="hidden" name="id[]" value="<?php echo $shuffle_id; ?>">
        <input type="hidden" name="username[]" value="<?php echo $shuffle_username; ?>">
        <input type="hidden" name="email[]" value="<?php echo $shuffle_email; ?>">

<?php 

    endforeach;
     if ( isset($_POST['shuffleButton'] ) ) :
        echo '<input type="submit" value="Finalize Draft Order" name="insert">';
    endif;

?>
</form>
<?php        
if (isset($_POST['insert'])) {
    $ordering_number = "1";

    $con = mysqli_connect("localhost", "", "", "");
    if (mysqli_connect_errno()) {
        printf("Connect failed: %s\n", mysqli_connect_error());
        exit();
    }
    $draft_stmt1 = $con->prepare("INSERT INTO user_players (user_id, firstname, lastname, username, email) VALUES (?, ?, ?, ?, ?)");
  if ( false===$draft_stmt1|| false===$draft_stmt2 ) {
         // Check Errors for prepare
        die('Add to user players prepare() failed: ' . htmlspecialchars($con->error));
    }
    $draft_stmt1->bind_param('issss', $shuffle_id, $shuffle_firstname, $shuffle_lastname, $shuffle_username, $shuffle_email);

    foreach ($_POST['id'] as $i => $shuffle_id) {
        $shuffle_firstname = $_POST['firstname'][$i];
        $shuffle_lastname = $_POST['lastname'][$i];
        $shuffle_username = $_POST['username'][$i];
        $shuffle_email = $_POST['email'][$i];

        $draft_stmt1->execute() or
            die('Add to user players execute() failed: ' . htmlspecialchars($draft_stmt1->error));

有谁知道为什么这不会按照我正在洗牌的顺序发送？我的数据库中有一个基本 ID 字段 table，这样每个插入的用户都会按照他们被起草的顺序获得 ID，但它不是那样工作的。

抱歉所有代码，我想确保不要忘记可能有用的东西谢谢！

更新

新代码

  $draft_stmt1 = $con->prepare("INSERT INTO user_players (id, user_id, firstname, lastname, username, email) VALUES (?, ?, ?, ?, ?, ?)");

 if ( false===$draft_stmt1|| false===$draft_stmt2 || false===$draft_stmt3) {
         // Check Errors for prepare
        die('Add to user players prepare() failed: ' . htmlspecialchars($con->error));
    }
    $draft_stmt1->bind_param('iissss', $i, $shuffle_id, $shuffle_firstname, $shuffle_lastname, $shuffle_username, $shuffle_email);

    foreach ($_POST['id'] as $i => $shuffle_id) {
        $shuffle_firstname = $_POST['firstname'][$i];
        $shuffle_lastname = $_POST['lastname'][$i];
        $shuffle_username = $_POST['username'][$i];
        $shuffle_email = $_POST['email'][$i];

        $draft_stmt1->execute() or
            die('Add to user players execute() failed: ' . htmlspecialchars($draft_stmt1->error));

这是用户被洗牌的地方..

$query = mysqli_query($con, "SELECT * FROM users WHERE group IN (3,4,5) ORDER BY id");
$array = array();

while ($row = mysqli_fetch_assoc($query)) { 
$array[] = array( 
'id' => $row['id'],
'firstname' => $row['firstname'], 
'lastname' => $row['lastname'], 
'username' => $row['username'], 
'email' => $row['email'] 
); 

if (isset($_POST['shuffle'])) {

   }
}
shuffle($array);
    echo json_encode($array);

Answer 1

向 table 添加一列 shuffle_order，并在该列中存储从表单中收到玩家的顺序。

$draft_stmt1 = $con->prepare("INSERT INTO user_players (shuffle_order, user_id, firstname, lastname, username, email) VALUES (?, ?, ?, ?, ?, ?)");
if ( false===$draft_stmt1|| false===$draft_stmt2 ) {
    // Check Errors for prepare
    die('Add to user players prepare() failed: ' . htmlspecialchars($con->error));
}
$draft_stmt1->bind_param('iissss', $i, $shuffle_id, $shuffle_firstname, $shuffle_lastname, $shuffle_username, $shuffle_email);

foreach ($_POST['id'] as $i => $shuffle_id) {
    $shuffle_firstname = $_POST['firstname'][$i];
    $shuffle_lastname = $_POST['lastname'][$i];
    $shuffle_username = $_POST['username'][$i];
    $shuffle_email = $_POST['email'][$i];

    $draft_stmt1->execute() or
        die('Add to user players execute() failed: ' . htmlspecialchars($draft_stmt1->error));
}

然后当你想按打乱顺序处理数据时，在SELECT查询中使用ORDER BY shuffle_order。

打乱后的结果未按打乱后的顺序发送到数据库

Shuffled results not sending to database in the shuffled order

php

mysql

shuffle