我在我的數據庫中有兩個表。當我執行下面的PHP腳本時,我收到錯誤。我認爲查詢是正確的,但我怎麼綁定參數錯誤?mysqli綁定參數的錯誤
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():綁定變量的數目不
匹配事先準備好的聲明字段數
修復:如果您正在從數據庫讀取圖片,請確保只返回查詢中的斑點數據。 (從SELECT語句中刪除cards.CARD_TYPE。)
的功能令'準備() - > bind_param() - >執行()' – diEcho 2012-07-28 05:22:55