Question

您好我正在尝试echo来自Product Table的{{1}}产品名称，而是echo来自Category table的名称。除了从2个表中的一个更改名称之外，它们是否可以阻止它？

Table: Product

   Pid     Name
    1      man
    2      woman
    3      child


Table: Category

  CatID    Name       Pid
    1       pen        1
    2       pen        1
    3       pencil     2
    3       red_pen    3


Table: active_product

  ATpid  size  pid   price
    1     12    1     10
    2     14    2     15
    3     16    3     20
    4     18    4     30

PDO：

$stmt = $conn->prepare("
    SELECT Product.Name, Category.Name, size, Price
    FROM active_Product, Product, Category
    WHERE Product.ProdID ='$item_id'
    AND size = '$size '
    AND Category.Name = '$pcategory'
    Limit  1");
$stmt->bindParam('$item_id',$item_id);
$stmt->bindParam('$size',$size); 
$stmt->bindParam('$pcategory',$pcategory); 
$stmt->execute();
foreach($stmt->fetchAll(PDO::FETCH_ASSOC) as $row ){
    $product_name = $row["Name"];

Answer 1

试试这个：

SELECT Product.Name AS PName, Category.Name AS CName, size, price 
FROM active_product, Product, Category 
WHERE Product.pid=1 AND size=10 AND Name.Category="pen" Limit 1

然后在PHP中，执行以下操作：

$product_name = $row["PName"];

所以你的最终产品应该是这样的：

$stmt = $conn->prepare("
    SELECT Product.Name AS PName, Category.Name AS CName, size, Price
    FROM active_Product, Product, Category
    WHERE Product.ProdID ='$item_id'
    AND size = '$size '
    AND Category.Name = '$pcategory'
    Limit  1");

...

foreach($stmt->fetchAll(PDO::FETCH_ASSOC) as $row ){
    $product_name = $row["PName"];

Answer 2

当您获取值时，PHP不会使用该表来限定名称。它将使用最后一个值，因此Name将引用类别名称。您可以撤消选择Product / Category名称的顺序，但使用别名更有意义。

SELECT Product.Name AS proName, Category.Name AS catName

$product_name = $row['proName'];

Answer 3

除了已经提到的更正之外，你没有绑定任何变量，你只是在sql语句中注入它们。

您应该使用：

$stmt = $conn->prepare("
SELECT Product.Name as productName, Category.Name as categoryName, size, Price
FROM active_Product, Product, Category
WHERE Product.ProdID =:item_id
AND size = :size  AND Category.Name = :pcategory Limit  1");
$stmt->bindParam(':item_id',$item_id);
$stmt->bindParam(':size',$size); 
$stmt->bindParam(':pcategory',$pcategory);

而不是：

$stmt = $conn->prepare("
SELECT Product.Name, Category.Name, size, Price
FROM active_Product, Product, Category
WHERE Product.ProdID ='$item_id'
AND size = '$size '  AND Category.Name = '$pcategory' Limit  1");
$stmt->bindParam('$item_id',$item_id);
$stmt->bindParam('$size',$size); 
$stmt->bindParam('$pcategory',$pcategory);

请注意，您需要使用:而不是$，并且不需要引用变量。

PHP回显错误的值

3 个答案: