php - mysqli绑定(bind)参数错误

Question

我的数据库里有两张表。当我执行下面的php脚本时，我得到了错误。我认为查询是正确的，但如何将参数绑定为错误的呢？

Display
+-------+------------+-------------+
| Index | DISPLAY_ID | Picture_ID  |
+-------+------------+-------------+
|     1 |         12 | longblob    |
+-------+------------+-------------+

和

Cards
+--------+------------+------------+-----------------+
| Card_ID| DISPLAY_ID | Card_Type  |  Card_name      |
+--------+------------+------------+-----------------+
|        |            |            |                 |
+--------+------------+---------=--+-----------------+

 <?php
$mysqli=mysqli_connect('localhost','root','','draftdb');
if (!$mysqli)
die("Can't connect to MySQL: ".mysqli_connect_error());


$stmt = $mysqli->prepare("SELECT cards.CARD_TYPE, display.PICTURE_ID 
FROM cards  
INNER JOIN display ON cards.DISPLAY_ID = display.DISPLAY_ID 
WHERE display.DISPLAY_ID=? AND cards.CARD_TYPE =?" );


$displayid=11;
$cardtype='Mythic';

$stmt->bind_param("si", $displayid, $cardtype);
$stmt->execute();
$stmt->bind_result($image);
$stmt->fetch();
//header("Content-Type: image/jpeg");
echo $image; 
?>

警告：mysqli_stmt：：bind_result（）：：绑定变量的数目与中准备的语句中的字段数目不匹配
修复：如果您正在从数据库读取图片，请确保只在查询中返回blob数据。（从SELECT语句中删除cards.CARD_TYPE。）

Answer 1

尝试以下两种修复方法：
摆脱：if (!$mysqli->query($stmt)) { printf("Errormessage: %s\n", $mysqli->error); }
调用execute()时执行查询
您需要绑定两个参数，然后传递两个但只列出一个类型。将bind_param调用更改为：
$stmt->bind_param("si", $cardtype, $displayid); // s for string, i for int
编辑：尝试将代码更改为：

$stmt = $mysqli->prepare("SELECT cards.CARD_TYPE, display.PICTURE_ID 
FROM cards  
INNER JOIN display ON cards.DISPLAY_ID = display.DISPLAY_ID 
WHERE display.DISPLAY_ID=? AND cards.CARD_TYPE =?" );


$displayid=11;
$cardtype='Mythic';

$stmt->bind_param("is", $displayid, $cardtype);
$stmt->execute();