多个 MySql 插入和更新未按预期运行

Question

我正在尝试将信息插入到一个表中，并在一个字符串中更新另一个表（在同一个数据库中）中的两行。我的印象是，您可以通过在每次尝试后添加一个分号来非常简单地做到这一点，它会在同一个字符串中单独操作它们。

我想不出任何其他方法来实现在“games”表中插入数据并同时更新“players”表中两个唯一行中的数据的预期效果。

这是我的代码...

<?php

include_once('database-details.php');

$winner         = mysqli_real_escape_string($db, $_REQUEST['winner']);
$loser              = mysqli_real_escape_string($db, $_REQUEST['loser']);
$balled         = mysqli_real_escape_string($db, $_REQUEST['balled']);
$fixtures           = 'games';
$table              = 'players';

$sql = "INSERT INTO `$fixtures` "
    . "(`id`, `winner`, `loser`, `balled`) VALUES "
    . "(NULL, '$winner', '$loser', '$balled') ;"
    . "UPDATE `$table` SET apps = apps + 1, wins = wins + 1, balled = balled + $balled WHERE id = $winner ;"
    . "UPDATE `$table` SET apps = apps + 1, losses = losses + 1, balled = balled - $balled WHERE id = $loser";

if(!$result = $db->query($sql)){
    die('There was an error running the query [' . $db->error . ']');
}

$response = (object)array('status' => 'error');

if($result){
    $response = (object)array(
        'status'    => 'ok',
        'id'        => $db->insert_id,
    );
}

header('Content-type: application/json');
echo json_encode($response);

?>

脚注... 表单使用ajax提交，然后更新前端的内容。我可以确认，当我测试过的请求中只有一个请求时，所有三个单独提交并且每个都运行良好时，提交工作正常。只有当我将它们组合起来时它才会中断。

提前感谢您的帮助。

Answer 1

您必须在同一个查询中执行更新查询吗？

也没有意义

$response = (object)array('status' => 'error');

if($result){

因为你已经停止了上面那行的脚本执行

if(!$result = $db->query($sql)){

不妨试试：

<?php

include_once('database-details.php');

$winner         = mysqli_real_escape_string($db, $_REQUEST['winner']);
$loser              = mysqli_real_escape_string($db, $_REQUEST['loser']);
$balled         = mysqli_real_escape_string($db, $_REQUEST['balled']);
$fixtures           = 'games';
$table              = 'players';

$sql = "INSERT INTO `$fixtures` "
    . "(`id`, `winner`, `loser`, `balled`) VALUES "
    . "(NULL, '$winner', '$loser', '$balled') ;"

if(!$result = $db->query($sql)){
    die('There was an error running the query [' . $db->error . ']');
else{
    $insertid = $db->insert_id
    $sql = "UPDATE `$table` SET apps = apps + 1, wins = wins + 1, balled = balled + $balled WHERE id = $winner ;"
    if(!$result = $db->query($sql)){
        die('There was an error running the query [' . $db->error . ']');
    }
    $sql = "UPDATE `$table` SET apps = apps + 1, losses = losses + 1, balled = balled - $balled WHERE id = $loser";
    if(!$result = $db->query($sql)){
        die('There was an error running the query [' . $db->error . ']');
    }
    $response = (object)array(
        'status'    => 'ok',
        'id'        => $insertid ,
    );
}

header('Content-type: application/json');
echo json_encode($response);

?>

根据您的 $db 类，如果您在一个查询中运行多个 sql 语句，则 insert_id 的值可能不是您所期望的。

Answer 2

这看起来像是一个并发问题。您可能想使用关闭自动提交，并使用 mysqli_begin_transaction() 设置块，执行您的查询和 mysqli_commit() 来处理块。

类似：

    mysqli_autocommit($db, FALSE);
    mysqli_begin_transaction();
    //now call query() as many times as needed
    mysqli_commit(); 

Answer 3

感谢大家的快速回复。最后，为 mysqli_multi_query() 提供的链接@AR 起到了作用。这是它现在的样子（如果有人感兴趣的话）......

<?php

include_once('database-details.php');

$winner    = mysqli_real_escape_string($db, $_REQUEST['winner']);
$loser     = mysqli_real_escape_string($db, $_REQUEST['loser']);
$balled    = mysqli_real_escape_string($db, $_REQUEST['balled']);
$fixtures  = 'games';
$table     = 'players';

$sql = "INSERT INTO `$fixtures` "
    . "(`id`, `winner`, `loser`, `balled`) VALUES "
    . "(NULL, '$winner', '$loser', '$balled') ;"
    . "UPDATE `$table` SET apps = apps + 1, wins = wins + 1, balled = balled + $balled WHERE id = $winner ;"
    . "UPDATE `$table` SET apps = apps + 1, losses = losses + 1, balled = balled + $balled WHERE id = $loser";

if (!$db->multi_query($sql)) {
    echo "Multi query failed: (" . $db->errno . ") " . $db->error;
}

do {
    if ($result = $db->store_result()) {
        $response = (object)array(
            'status'    => 'ok'
        );
        $result->free();
    }
} while ($db->more_results() && $db->next_result());

header('Content-type: application/json');
echo json_encode($response);

?>    

PS - 感谢@Hadyn Dickson 对查询的建议。