Question

我是陈述的新手，所以请慢慢来找我。我检查了其他人提出的问题，但没有看到解决我问题的解决方案。

我正在尝试使用预准备语句创建一个用户页面，以便他们可以将产品添加到他们的商店。我想从商店获取store_id并在插入产品表单中插入produtcs。

我尝试了几种方法，但它们没有用。以下是我的尝试：

连接

$mysqli = new mysqli(host, dbase, username, password);

第一种方法准备语句：我也试过这种方法而没有bind_result。

if ($stmt = $mysqli->prepare("INSERT INTO products (user_id, cat_id, store_id, item_name, item_code, item_description, item_qtty, item_price, item_seo_url, item_image, item_date) SELECT store_id FROM stores WHERE user_id = ?")) {
    $stmt->bind_param("i", $user_id);
    $user_id = filterString($_SESSION['id']);
    $stmt->execute();
        $stmt->bind_result($store_id);
        if($stmt->fetch()){
        echo " Records created successfully. Redirect to landing page";
        header("location: index.php");
        exit();
    } else{
        echo "Something went wrong. Please try again later.";
       }
}
$stmt->close();

第二个方法sql prepare语句：我也试过这个但是没有用：

    $sql = "INSERT INTO products (user_id, cat_id, store_id, item_name, item_code, item_description, item_qtty, item_price, item_seo_url, item_image, item_date) SELECT ?, store_id FROM stores WHERE user_id = ?";
    if($stmt = $mysqli->prepare($sql)){
    $stmt->bind_param("iiisiisiisss", $user_id, $cat_id, $store_id, $item_name, $item_code, $item_description, $item_qtty, $item_price, $item_seo_url, $item_image, $item_date);
    $user_id = $_SESSION['id'];
    $cat_id = $cat_id;
    $store_id = $store_id;
    $item_name = $item_name;
    $item_code = $item_code;
    $item_description = $item_description;
    $item_qtty = $item_qtty;
    $item_price = $item_price;
    $store_seo_url = seo_url($item_name);
    $item_image = $vtyol;
    $item_date = $date;
    if($stmt->execute()){
        echo " Records created successfully. Redirect to landing page";
        header("location: index.php");
        exit();
    } else{
        echo "Something went wrong. Please try again later.";
    }
}
$stmt->close();

没有机会从商店获取store_id，我在页面中回显商店ID，看看我是否得到它，它是空的。我怎样才能让它发挥作用？我是否需要在第一个方法中声明bind_param中的所有值？（我试过并没有工作）。如果是这样，如何添加条款$stmt->bind_param("i", $user_id);。

我真的不知道还有什么可以尝试，需要你的建议和帮助。到医院的Gtg现在可以在1小时内回来，会回答你的问题和答案。

谢谢大家

最后一个例子我尝试使用来自@Michael Eugene Yuen的代码，因为无法从商店表中获取$ store_id，所以一直说错了。我的代码很长，所以我缩短了它们并尝试得到相同的结果。

这是最后一个不动摇的例子：

$sql = "INSERT INTO products (
user_id, store_id, name, salary
)  
 SELECT ?, ?, ?, ?,
`store_id` FROM stores WHERE user_id = ?";
$stmt = $mysqli->prepare($sql);
    $stmt->bind_param("iisii", $user_id, $store_id, $name, $salary, $user_id);
    $user_id = $user_id;
    $store_id = $store_id;
    $name = $name;
    $salary = $salary;
    if($stmt->execute()){
        header("location: index.php");
        exit();
    } else{
        echo "Something went wrong. Please try again later.";
    }
$stmt->close();

两个表的数据库打击：

CREATE TABLE IF NOT EXISTS `products` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `user_id` varchar(100) NOT NULL,
  `store_id` varchar(100) NOT NULL,
  `name` varchar(255) NOT NULL,
  `salary` int(10) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1;

CREATE TABLE IF NOT EXISTS `stores` (
  `store_id` int(11) NOT NULL AUTO_INCREMENT,
  `user_id` varchar(100) NOT NULL,
  `name` varchar(255) NOT NULL,
  `salary` int(10) NOT NULL,
  PRIMARY KEY (`store_id`)
) ENGINE=MyISAM AUTO_INCREMENT=3 DEFAULT CHARSET=latin1;

INSERT INTO `stores` (`store_id`, `user_id`, `name`, `salary`) VALUES
(1, '12', 'aaaaaaaaaaad', 12),
(2, '12', 'sada', 1234656);

Answer 1

您试图将数据插入11列，但只有1个值传递到您的语句

您的陈述应如下所示：

$sql = "INSERT INTO products (
            user_id, cat_id, item_name, item_code, item_description,
            item_qtty, item_price, item_seo_url, item_image, item_date,
            store_id
                )
        SELECT ?, ?, ?, ?, ?,
               ?, ?, ?, ?, ?,
               `store_id` FROM stores WHERE user_id = ?";

$stmt = $mysqli->prepare($sql);

将$ user_id绑定两次。第一个是您的第一个占位符，最后一个是您的子选择语句（http://ecomgx17.ecomtoday.com/edi/）

$stmt->bind_param("iisisiisssi", $user_id, $cat_id, $item_name, $item_code, $item_description, $item_qtty, $item_price, $item_seo_url, $item_image, $item_date, $user_id);

$user_id = $_SESSION['id'];
$cat_id = $cat_id;
$item_name = $item_name;
$item_code = $item_code;
$item_description = $item_description;
$item_qtty = $item_qtty;
$item_price = $item_price;
$store_seo_url = seo_url($item_name);
$item_image = $vtyol;
$item_date = $date;

if($stmt->execute()){
        echo " Records created successfully. Redirect to landing page";
        $stmt->close();         
        header("location: index.php");
        exit();
} else{
        echo "Something went wrong. Please try again later.";
}

根据您新添加的示例2 ，列数，占位符数再次与bind_param数不匹配。此外，当您查看商店表时，您在同一user_id下有两条记录。那么您希望将哪个store_id插入商店表？

我现在使用LIMIT 1但当然这不是正确的做法。

$sql = "INSERT INTO products (
            user_id, name, salary, store_id
        )  
        SELECT ?, ?, ?, `store_id` FROM stores WHERE user_id = ? LIMIT 1";
$stmt = $mysqli->prepare($sql);
$stmt->bind_param("isii", $user_id, $name, $salary, $user_id);
$user_id = 12;
$name = 'john';
$salary = 1200;
if($stmt->execute()){
    header("location: index.php");
    exit();
} else {
    echo "Something went wrong. Please try again later.";
}
$stmt->close();

这是一个带有硬编码值的sql语句：

$sql = "INSERT INTO products (
            user_id, name, salary, store_id
        )
        SELECT 12, 'john', 1200,
        `store_id` FROM stores WHERE user_id = 12 LIMIT 1";

这与变量相同：

$sql = "INSERT INTO products (
           user_id, name, salary, store_id
        )
        SELECT $user_id, $name, $salary,
        `store_id` FROM stores WHERE user_id = $user_id LIMIT 1";

这与占位符相同：

$sql = "INSERT INTO products (
        user_id, name, salary, store_id
        )  
        SELECT ?, ?, ?,
        `store_id` FROM stores WHERE user_id = ? LIMIT 1";

$stmt->bind_param("isii", $user_id, $name, $salary, $user_id);
$user_id = 12;
$name = 'john';
$salary = 1200;

如何从商店获取值并插入到产品表中？

1 个答案: