php - MySQLi 准备好的语句 - 不同步的命令仍然返回结果

Question

我要么失去了情节，要么有人在玩弄我的思想:/

我已将此代码与我的应用程序的其余部分分开以尝试和调试，还硬编码了 $guideid 以便于阅读。

我有以下代码(此脚本中没有其他内容作为测试，即没有其他查询):

     <?php
     define("DB_HOST", "127.0.0.1");
     define("DB_NAME", "xxxxx");
     define("DB_USER", "xxxxx");
     define("DB_PASS", "xxxxx");

     $mysqli = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);

     $stmt = $mysqli->prepare("SELECT `guides_listings`.`listing_id`, `guide_slug`, `guide_name_en`, listing_name, `listing_slug`, `slogo`.`filename` AS `slogoname`, `hlogo`.`filename` AS `hlogoname`, `vlogo`.`filename` AS `vlogoname`

                                      FROM `guides_listings`
                                      JOIN `guides` ON `guides_listings`.`listing_guide` = `guides`.`guide_id`
                                      LEFT JOIN `guides_listings_pics` AS `slogo`
                                        ON `slogo`.`listing_id` = `guides_listings`.`listing_id`
                                        AND `slogo`.`type` = 'slogo'
                                      LEFT JOIN `guides_listings_pics` AS `hlogo`
                                        ON `hlogo`.`listing_id` = `guides_listings`.`listing_id`
                                        AND `hlogo`.`type` = 'hlogo'
                                      LEFT JOIN `guides_listings_pics` AS `vlogo`
                                        ON `vlogo`.`listing_id` = `guides_listings`.`listing_id`
                                        AND `vlogo`.`type` = 'vlogo'
                                      WHERE (`slogo`.`filename` IS NOT NULL OR `hlogo`.`filename` IS NOT NULL OR `vlogo`.`filename` IS NOT NULL)
                                        AND `guides_listings`.`listing_guide` = ?
                                      GROUP BY `guides_listings`.`listing_id`
                                      ORDER BY RAND() 
                                      LIMIT 12");

    $stmt->bind_param("i",$guideid);
    $guideid = 2;   

    $stmt->execute();       

    $stmt->bind_result($listing_id,$guide_slug,$guide_name,$listing_name,$listing_slug,$slogo,$hlogo,$vlogo);           

    while($stmt->fetch()) {
        $results->data[] = array('listing_id'=>$listing_id,'guide_slug'=>$guide_slug,'guide_name'=>$guide_name,'listing_name'=>$listing_name,'listing_slug'=>$listing_slug,'slogo'=>$slogo,'hlogo'=>$hlogo,'vlogo'=>$vlogo);
    }

    $stmt->close();
    ?>

这导致命令不同步；你现在不能在 execute() 之后运行这个命令

但是仍然返回正确的结果。

我是否遗漏了一些明显的东西，说明为什么它仍然有效但会发出此错误？

编辑

我刚刚进行了一些测试，这似乎发生在所有准备好的查询上，请参见下面的示例:

$mysqli = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);

$stmt = $mysqli->stmt_init();
$stmt->prepare("SELECT `listing_id` FROM `guides_listings` LIMIT 1");
$stmt->execute(); 

print_r($mysqli);  

$stmt->bind_result($listing_id);           

$stmt->fetch();
$results->data[] = array('listing_id'=>$listing_id);
$stmt->close();

print_r 结果如下:

mysqli Object
(
[affected_rows] => -1
[client_info] => 5.6.21
[client_version] => 50621
[connect_errno] => 0
[connect_error] => 
[errno] => 0
[error] => 
[error_list] => Array
    (
    )

[field_count] => 1
[host_info] => 127.0.0.1 via TCP/IP
[info] => 
[insert_id] => 0
[server_info] => 5.6.21
[server_version] => 50621
[stat] => Commands out of sync; you can't run this command now
[sqlstate] => HY000
[protocol_version] => 10
[thread_id] => 19462371
[warning_count] => 0
)

也许我真的在看这里的错误？

Answer 1

• 在绑定(bind)之前分配$guideid。

• 尝试在 execute() 和 bind_result() 之间调用 $stmt->store_result()。

• 看起来 $stmt 并没有实际定义。通常我会做 $stmt = mysqli->prepare(...)