Question

我有两张桌子：

table {node}
`nid`, `uid`, `type`
 1    1   basketball
 2    1   basketball
 3    1   football
 4    2   football
 5    2   basketball

table {strato_ticket}
`tid`, `author_uid`, `purpose`, `active`
 1      1   'Ticket to a basketball game' TRUE
 2      1   'Ticket to a football game'   TRUE
 3      2   'Ticket to a football game'   FALSE

我想生成一个报告，计算每种节点的数量，然后计算每个用户与该类节点关联的活动票证数。

我的解决方案使用SQL和PHP的组合：我对我感兴趣的每种节点都有一个PHP循环，这简化了SQL查询，并从“类型”转换为“目的”，例如

$node_types = array('basketball', 'football');
foreach($node_types as $node){
  switch($type){
    case 'basketball':
      $purpose = array('Ticket to a basketball node');
      break;
    case 'football':
      $purpose = array('Ticket to a football game');
      break;
  }
  $where = " WHERE ({strato_ticket}.`purpose` = '"
    .implode("' OR {strato_ticket}.`purpose` = '",$purpose)."')";

最后我有麻烦的地方，SQL查询。当我只计算每个用户拥有的节点时，它运行良好：

$query = "
      SELECT uid, count( * ) AS nodes_owned
      FROM {node} WHERE `type` = '$type'
      GROUP BY uid ORDER BY nodes_owned DESC
      ";
  $query = db_query($query);

output:
Now displaying info for basketball.
uid nodes_owned
 1       2
 2       1
Now displaying info for football.
uid nodes_owned
 1       1
 2       1

但是现在我需要查询另一个表strato_ticket，事情变得复杂，我的查询返回FALSE而不会抛出错误（我认为）。

 $query = "
    SELECT count(*) as tickets
    FROM {strato_ticket} INNER JOIN (
        SELECT node.uid, count( * ) AS nodes_owned
        FROM {node} WHERE `type` = '$type'
        GROUP BY uid
      ) AS {nodecount}
      ON {strato_ticket}.`author_uid` = {nodecount}.`uid`
      $where
      GROUP BY nodecount.uid ORDER BY nodecount.nodes_owned DESC
      ";
  $query = db_query($query);

我对SQL不是很了解，我不太确定它是如何破坏的。可以用一些帮助吗？

理想情况下希望看到

uid nodes_owned tickets
//basketball
 1       2        1
 2       1        0
//football 
 1       1        1
 2       1        0

Answer 1

除了占位符，我可以稍后介绍，我认为这解决了它。

$form = array();
$node_types = array('basketball','football');
// if($user->uid == 1){
  $form[$type][] = array('#value'=>"Showing how many of each node type each user owns.".'<br/>');
  foreach($node_types as $type){
  // Count the number of nodes each user owns of $type.
  $form[$type][] = array('#value'=>"Now displaying info for $type".'s. <br/>');
  switch($type){
    case 'basketball':
      $purpose = array('ticket to a basketball game', 'basketball');
    break;
    case 'football':
      $purpose = array('ticket to a football game');
    break;
  }
  $purpose = implode("', '", $purpose);
  //@todo : Make a temporary table to query against so I'm not hitting node table multiple times.
  $ticketquery = "
    SELECT author_uid, purpose, COUNT( * ) AS invitees_accepted
    FROM {strato_ticket}
    WHERE purpose IN ('$purpose')
    GROUP BY author_uid, `purpose`
  ";
  $nodequery = "
    SELECT node.uid, count( * ) AS nodes_owned, type
    FROM {node}
    WHERE `type` IN ('$type')
    GROUP BY uid, type";
  $query = "
    SELECT * FROM
    ($nodequery) AS nt
    JOIN
    ($ticketquery) AS tt
    ON nt.uid = tt.author_uid
    GROUP BY nt.uid ORDER BY nt.nodes_owned DESC
   ";

  drupal_set_message('Query is <br/>'.$query);
  //return;
  $query = db_query($query);
  $first = true;
  while ($rec = db_fetch_object($query)){
    if($first){
      $form[$type][] = array('#value'=>"And the winner is: ".print_r($rec, true).'<br/>');
      $first = false;
    }
    else {
     $form[$type][] = array('#value'=>print_r($rec, true).'<br/>');
    }
  }
 // }
}

通过PHP进行基于Drupal的SQL查询：count t1 join t2

1 个答案: