Question

所以我试图学习在我的查询中使用预处理语句，因为它使用了旧的mysql方式，但我没有太多运气。

这是我的代码

<?php
$stmt = $db->stmt_init();

if($stmt->prepare("SELECT l.listingID, l.title, l.description, l.dateListed
FROM tbl_listings AS l
LEFT JOIN tbl_listing_type AS lt ON lt.listingID = l.listingID
LEFT JOIN tbl_type AS t ON t.typeID = lt.typeID
WHERE lt.typeID =?"))
{
    $stmt->bind_param("i",$type);
    $type = 1;
    $stmt->bind_result($id, $title, $description, $date);

    while($stmt->fetch())
    {
        echo $id . ' - ' . $title . ' - ' . $description . ' - '.$date."<br />";    
    }
    $stmt->close();
}
else
{
    echo "error";
}

但它不打印任何东西，我在我的phpmyadmin中运行查询而不是？它返回记录，所以我知道查询是正确的，但我不确定我正确使用准备好的位？有人可以告诉我我可能出错的地方吗？

非常感谢

Answer 1

您需要在绑定之前将值分配给$type：

$type = 1;
$stmt->bind_param("i",$type);

Answer 2

$stmt = $db->stmt_init();

if($stmt->prepare("SELECT l.listingID, l.title, l.description, l.dateListed
FROM tbl_listings AS l
LEFT JOIN tbl_listing_type AS lt ON lt.listingID = l.listingID
LEFT JOIN tbl_type AS t ON t.typeID = lt.typeID
WHERE lt.typeID =?"))
{
    $type = 1;
    $stmt->bind_param("i",$type);
    $stmt->execute(); <-- Was missing this in my original code
    $stmt->bind_result($id, $title, $description, $date);

    while($stmt->fetch())
    {
        echo $id . ' - ' . $title . ' - ' . $description . ' - '. $date . "<br />"; 
    }
    $stmt->close();
}?>

让它工作:)缺少执行查询所需的一行代码。

试图学习PHP编写的SQL语句，但我做错了但我不确定是什么？

2 个答案: