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我无法让我的存储过程或视图从数据库中检索内容。我不知道我的存储过程或视图是否有效，但我知道没有处理这些工作的 php。谢谢，麻烦您了！

下面是我的视图和存储过程。 

CREATE VIEW viewActRole AS
SELECT `tblMovie`.`MovieName`, `tblActor`.`ActFirstName`, `tblActor`.`ActLastName`,    `tblRole`.`Role` 
FROM `tblRole`, `tblActor`, `tblMovie` 
WHERE `tblRole`.`ActID` = `tblActor`.`ActID` AND `tblRole`.`MovieID`=`tblMovie`.`MovieID`;


DELIMITER $$

CREATE PROCEDURE `lookUpMovie`(IN `actorNameVAR` INT)
BEGIN
SELECT * FROM `viewActRole` WHERE `viewActRole`.`MovieID` = `actorNameVAR`;
END$$

DELIMITER ;

下面是我的php代码。我知道在添加存储过程之前一切正常。

<?php
$mysqli = new mysqli("BLAH", "BLAH", "BLAH", "BLAH");

$dd = $_POST['displaydropdown'];

// used fixed-width font
echo "<pre>\n";
$sp = "CALL lookUpMovie(".$dd.")";
// Get person likes for
echo "$sp";
$r1 = $mysqli->query($sp);

$r2 = mysql_query("SELECT C.MovieName, B.ActFirstName, B.ActLastName, A.Role 
    FROM tblRole A, tblActor B, tblMovie C WHERE A.ActID = B.ActID AND A.MovieID=C.MovieID AND A.MovieID = '$dd'");
$result2 = mysql_fetch_array($r2);

echo "Albums of Artist: $result2[0]\n";
$fmt = "%-20s %-20s %-20s %-20s \n";
printf($fmt, "Movie Name", "Actor First Name", "Actor Last Name", "Role");

// loop over courses, printing each one
while ($result = mysql_fetch_array($r1))
{ 
    printf($fmt,$result[0], $result[1], $result[2], $result[3]);
}

// free result set
mysql_free_result($r1);

echo "</pre>\n";
?>
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1 回答 1

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看起来您正在混合和匹配 mysqli 面向对象的调用与已弃用的 mysql_* 函数。我不确定你能做到这一点。

你正在做：

$r1 = $mysqli->query($sp)

然后将其传递给

mysql_fetch_assoc($r1);

循环遍历您的结果，您需要执行以下操作：

while ($result = $r1->fetch_array())
{ 
    printf($fmt,$result[0], $result[1], $result[2], $result[3]);
}
于 2013-06-09T03:12:20.847 回答