順序按類別

Question

結果，我想：

campaign 1 
    offer 1.1 
    offer 1.2 
    offer 1.3 
    (and so forth) 

campaign 2 
    offer 2.1 
    offer 2.2 
    offer 2.3 
    (and so forth) 

這些都是我的表：

campaign (tablename) 
    campaign_id (used) 
    campaign_name 

offer (tablename) 
    offer_id 
    offer_parentid (uses category_id find which category the item is supposed to be paired 
with.) 
    offer_name 

我有使用兩個表的小經驗，但我覺得它的這種東西是有用的。 我的問題是我已經嘗試了PHP的每一個方面，我知道，我不能得到我想要的結果。 這是我幾個小時的鮮血，汗水和淚水後的最終結果是：

$get_campaign = mysql_query("SELECT * FROM campaign ORDER BY campaign_id DESC") 
    or die(mysql_error()); 

$get_offers = mysql_query("SELECT * FROM offers JOIN campaign ON  offers.offer_parentid=campaign.campaign_id ORDER BY campaign_id DESC ") 
    or die(mysql_error()); 

while($row = mysql_fetch_array($get_campaign)){ 
echo $row['campaign_name']."</br>"; 
while($row = mysql_fetch_assoc($get_offers)){ 
    echo $row['offer_name']."</br>"; 
    } 
} 

結果變成：

Campaign 1 
offer 1.1 
offer 1.2 
offer 2.1 
offer 2.2 
campaign 2 

感謝所有幫助我能得到

Answer 1

你需要讓你的SQL for your second query conditional on the first query：

$get_campaign = mysql_query("SELECT * FROM campaign ORDER BY campaign_id DESC") 
    or die(mysql_error()); 


while($row = mysql_fetch_array($get_campaign)){ 
    echo $row['campaign_name']."</br>"; 
    $get_offers = mysql_query("SELECT * FROM offers WHERE campaign_id = " . $row['campaign_id'] . " ORDER BY campaign_id DESC ") 
    or die(mysql_error()); 

    while($offerRow = mysql_fetch_assoc($get_offers)){ 
     echo $offerRow['offer_name']."</br>"; 
    } 
} 

不是我假設您的廣告系列表中的主鍵列是「ID」。你也可以用一個連接在一個查詢中做到這一點，只是有一個循環，但取決於數據的大小和你真正從第一個表返回的數量，兩個循環對於性能來說更可取。

Answer 2

您可以使用一個查詢：

$query = 'SELECT c.campaign_name, o.offer_name 
FROM campaign c 
    INNER JOIN offers o ON (o.offer_parentid = c.campaign_id) 
ORDER BY c.campaign_id'; 

現在遍歷結果：

$campaign = ''; 
while($row = mysql_fetch_array($query)){ 
    // only display the campaign name when its not $campaign 
    if ($row['campaign_name'] != $campaign){ 
    echo $row['campaign_name']."</br>"; 
    $campaign = $row['campaign_name']; 
    } 
    echo $row['offer_name']."</br>"; 
} 

注：我必須說，你用choosen不佳列名。如果您參考商品表中的廣告系列表，則應選擇諸如offers.campaign_id之類的內容。原因在於你現在知道你需要什麼，但是當你獲得更多的表和列時，很難在表中選擇正確的列連接而不先查看你的PHP代碼。它可以節省您（和其他與您的代碼一起工作的人）很多時間。