2013-07-09 37 views
1

我有兩個表mysqli的選擇,而不是如果

 Products 
---------------------- 
ProductId | ProdutName 
    1  | test 
    2  | test 2 

    SubProducts 
----------------------------------------- 
SubProductId | SubProductName | ProductId 
     1  | sub_test |  1 
     2  | sub_test2 |  1 
     3  | sub_test3 |  1 
     4  | sub_test4 |  2 

我想從子產品

得到產品表中的每個產品編號都行,只有第一排因爲我想要這個:

ProductId = 1 
ProductName = test 
SubProductId = 1 
SubProductName = sub_test 

ProductId = 2 
ProductName = test2 
SubProductId = 4 
SubProductName = sub_test4 

我用if做了這個,但是我知道這是錯誤的,那麼可以用一個查詢來完成嗎?

我得到這個至今:

stmt = $connection->stmt_init(); 

if ($stmt->prepare(" 
    SELECT DISTINCT a.`ProductId`, a.`ProductName`, b.`SubProductId`, b.`SubProductName` 
    FROM `Products` as a, `SubProducts` as b 
    WHERE b.ProductId=a.ProductId 
")) 
{ 

    $stmt->execute(); 
    $stmt->bind_result($product_id, $product_name, $sub_product_id, $sub_product_name); 

    while($stmt->fetch()) 
    { 
     if ($current_id!=$product_id) 
     { 
     echo 'id = '.$product_id; 
     echo 'thumb = '.$sub_product_name; 

     $current_id = $product_id; 
     } 
    } 

    $stmt->close(); 
} 

回答

-1
SELECT a.`ProductId`, a.`ProductName`, b.`SubProductId`, b.`SubProductName` 
FROM `Products` as a, `SubProducts` as b 
WHERE 
b.`ProductId`=a.`ProductId` 
group by a.`ProductId` 

檢查鏈接SQL小提琴http://sqlfiddle.com/#!2/71778f/1/0

+0

OK,這是一個misstype,ü可以幫我查詢? – user2564276

+0

@ user2564276檢查SQL小提琴鏈接 –

+1

它工作!多謝! :) – user2564276