如何顯示SQL表中的所有記錄並將這些記錄與另一個表進行匹配？

Question

我已經將PHP變量$ accountnumber設置爲查看其個人檔案頁面的用戶。在頁面上，我用數據庫填充了用戶信息的一個塊，並且我有一個我們擁有的所有產品的列表，並且我希望在每個客戶旁邊放置一個複選標記，方法是分配一個類給它。

這裏是我的表：

products 
id | name | url | weight 
100 p1  p1.html 1 
101 p2  p2.html 2 
102 p3  p3.html 3 
103 p4  p4.html 4 
104 p5  p5.html 5 
105 p6  p6.html 6 

products_accounts 
account_number | product_id 
0000001   100 
0000001   104 
0000001   105 
0000002   101 
0000002   103 
0000002   104 
0000002   105 
0000003   100 
0000003   102 

我嘗試了LEFT OUTER JOIN，但無法確定$ ACCOUNTNUMBER匹配的products_accounts表的ACCOUNT_NUMBER特定的product_id。我能夠做到這一點的唯一方法是添加一個WHERE語句是這樣的：

WHERE products_acccounts.account_number = '$accountnumber' 

它給適當類的產品，但只顯示出他們，而不是所有的產品。

這裏是我的代碼：

$sql =" 
SELECT 
    products.id, 
    products.name, 
    products.url, 
    products_accounts.account_number 
FROM 
    products 
LEFT OUTER JOIN 
    products_accounts 
ON 
    products.id = products_accounts.product_id 

"; 

$sql .=" 
GROUP BY 
    products.id 
ORDER BY 
    products.weight 
"; 

$result = mysql_query($sql); 
while($row = mysql_fetch_array($result)) { 
    echo '<span class="'; if($row['account_number'] == '$accountnumber') 
    { echo'product_yes">'; } else { echo 'product_no">'; } 
    echo '<a href="' . $row['url'] . '">' . $row['name'] . '</a><br /></span>'; 
} 

如果客戶有不同的P2和P5所有的產品，它應該顯示是這樣的：

✓P1

P2 

✓P3 

✓P4 

P5 

✓P6 

Answer 1

$getproducts = mysql_query(" 
SELECT id, name, url 
FROM products 
ORDER BY weight ASC"); 

while ($rowproducts = mysql_fetch_assoc($getproducts)) { 

$product_id = $rowproduct['id']; 
$product_name = $rowproduct['name']; 
$product_url = $rowproduct['url']; 

$getuserhasproduct = mysql_query(" 
SELECT DISTINCT product_id 
FROM products_accounts 
WHERE account_number = $accountnumber 
AND product_id = $product_id"); 
$user_has_product = mysql_num_rows($getuserhasproduct); 

if($user_has_product){ 
$class = "checked"; 
} 

echo "<span class='$class'><a href='$product_url'>$product_name</a></span>"; 
unset($class); 
} // end loop 

這可能與性能幫助

$getproducts = mysql_query("SELECT id, name, url, 
(SELECT DISTINCT product_id 
FROM products_accounts 
WHERE account_number = '$accountnumber' 
AND product_id = products.id) AS product_count 
FROM products 
ORDER BY weight ASC"); 

while ($rowproducts = mysql_fetch_assoc($getproducts)) { 

$product_id = $rowproduct['id']; 
$product_name = $rowproduct['name']; 
$product_url = $rowproduct['url']; 
$product_count = $rowproduct['product_count']; 

if($product_count > 0){ 
$class = "checked"; 
} 

echo "<span class='$class'><a href='$product_url'>$product_name</a></span>"; 
unset($class); 
} // end loop 

Answer 2

你如果您想要檢索所有的內容，則嘗試在上下文中使用GROUP BY記錄。 GROUP BY子句只應用於如果要彙總數據（即獲得一堆記錄的總和，平均值等）。

Answer 3

SELECT 
    products.id, 
    products.name, 
    products.url, 
    products_accounts.account_number 
FROM 
    products 
LEFT OUTER JOIN 
    (SELECT * FROM products_accounts WHERE account_number = $account_number) as products 
ON 
    products.id = products_accounts.product_id 
WHERE 
"; 

$sql .=" 
GROUP BY 
    products.id 
ORDER BY 
    products.weight 
"; 

我認爲這是你的答案，你需要在連接之前過濾你的連接表。請檢查語法，因爲我不熟悉php。

Answer 4

這是更好地過濾掉使用SQL比PHP行，象下面這樣：

$sql =" 
SELECT 
    p.id, 
    p.name, 
    p.url, 
    pa.account_number 
FROM 
    products p 
LEFT OUTER JOIN 
    products_accounts pa 
ON 
    p.id = pa.product_id 
    AND 
    pa.account_number = ".mysql_real_escape_string($accountnumber)." 
ORDER BY 
    p.weight 
"; 


$result = mysql_query($sql); 
while($row = mysql_fetch_array($result)) { 
    echo '<span class="'; if(!is_null($row['account_number'])) 
    { echo'product_yes">'; } else { echo 'product_no">'; } 
    echo '<a href="' . $row['url'] . '">' . $row['name'] . '</a><br /></span>'; 
}