检查其他表中是否存在id

时间:2014-09-24 11:42:25

标签: php mysql

我想检查购物车表中是否存在product_id,是否存在定义变量$class = 'show'

这是我的代码:

$article = $db->query("
    SELECT 
        b.price as price,
        b.shop_id as shop_id,
        c.image as image,
        c.name as name,
        a.id as id
    FROM products AS a
    INNER JOIN prices AS b ON a.ean = b.sku
    INNER JOIN shops as c on b.shop_id = c.id
    WHERE a.id = '$product_id'
");     
while ($data = $article->fetch_object()) {
    $price = $data->price;
    $price2 = $data->price;
    $price = number_format($price, 2, ',', ''); 
    $shop_id = $data->shop_id;
    $logo = $data->image;
    $name = $data->name;

//some html code here

}

现在我不知道如何检查购物车表中是否存在该产品。在两个表中,uniqe是product_id

3 个答案:

答案 0 :(得分:2)

我为此目的使用LEFT JOIN(请注意CASE WHEN .. THEN ..部分)。它与cart表连接并选择一个名为exists_in_cart的列,如果产品存在于购物车中,则值为1,否则为0):

$article = $db->query("
    SELECT 
        b.price as price,
        b.shop_id as shop_id,
        c.image as image,
        c.name as name,
        a.id as id,
        CASE WHEN cart.product_id IS NULL THEN 0 ELSE 1 END as exists_in_cart
    FROM products AS a
    INNER JOIN prices AS b ON a.ean = b.sku
    INNER JOIN shops as c on b.shop_id = c.id
    LEFT JOIN cart on a.id = cart.product_id
    WHERE a.id = '$product_id'
");

如果产品存在于购物车中,您将获得1或者否则为0:

$productInCart = $data->exists_in_cart == 1 ? TRUE : FALSE;
if ($productInCart) {
   $class = 'show';
}
else {
   // ...
}

答案 1 :(得分:1)

检查是否返回了任何行,如果有,请设置变量。

if($article->num_rows) {
   $class = "show";
} 

或者,检查属性是否为空。

if( is_null($data->name) == FALSE ) {
   $class = "show";
}

答案 2 :(得分:0)

SELECT 
    b.price as price,
    b.shop_id as shop_id,
    c.image as image,
    c.name as name,
    a.id as id,
    CASE WHEN cart.product_id IS NULL THEN 0 ELSE 1 END as exists_in_cart
FROM products AS a
INNER JOIN prices AS b ON a.ean = b.sku
INNER JOIN shops as c on b.shop_id = c.id
LEFT JOIN cart on a.id = cart.product_id
WHERE a.id = '$product_id'

如果表中不存在该产品,则此查询将返回num_rows = 0,您已选择仅现有产品。