Question

我想要的结果：

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)

这些是我的表格：

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

我对使用两张桌子的经验不多，但我发现它对这种事情很有用。我的问题是我已经尝试了我知道的PHP的每个方面，我无法得到我想要的结果。这是我经过几个小时的血，汗和眼泪后的最终结果：

$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>";
    }
}

结果变为：

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

感谢您提供任何帮助

Answer 1

您需要以第一个查询条件为第二个查询生成SQL：

$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>";
    }
}

请注意，我假设您的广告系列表中的主键列是“id”。您也可以在一个带有连接的查询中执行此操作并且只有一个循环，但是根据数据的大小和您从第一个表中真正返回的数量，两个循环更适合性能。

Answer 2

您可以使用一个查询：

$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';

现在循环搜索结果：

$campaign = '';
while($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>";
}

注意：我必须说你使用选择不当的列名。如果您参考商品表中的广告系列表，则应选择offers.campaign_id。原因是你现在知道你需要什么，但是当你得到更多的表和列时，很难在没有首先查看PHP代码的情况下选择表之间的正确列连接。它可以节省你（以及其他使用你的代码的人）很多时间。

按类别排序

2 个答案: