使用JOIN參數轉換fql查詢sql

Question

我在這裏有此代碼，它允許我提取喜歡特定頁面的朋友，但我需要提取不喜歡某個頁面的朋友。

下面是代碼：

$users = $facebook->api(array('method' => 'fql.query', 
        'query' => "SELECT uid, first_name, pic_small FROM user WHERE uid IN(
    SELECT uid FROM page_fan WHERE page_id='411133928921438' AND uid IN (
        SELECT uid2 FROM friend WHERE uid1 = me() 
   ))")); 

我應該怎樣使用OUTER來得到這樣的加入或SMTH：

$users = $facebook->api(array('method' => 'fql.query', 
      'query' => "SELECT uid, first_name, pic_small FROM user WHERE uid NOT IN(
     SELECT uid FROM page_fan WHERE page_id='411133928921438' AND uid IN (
      SELECT uid2 FROM friend WHERE uid1 = me() 
     ))")); 

Answer 1

看到它不支持或OUTER JOIN NOT IN ，嘗試這樣做（不知道這是否是最好的方式，但它應該工作）

$likes = $facebook->api(array('method' => 'fql.query', 
    'query' => "SELECT uid, first_name, pic_small FROM user WHERE uid IN(
SELECT uid FROM page_fan WHERE page_id='411133928921438' AND uid IN (
    SELECT uid2 FROM friend WHERE uid1 = me() 
))")); 

$allfriends = $facebook->api(array('method' => 'fql.query', 
    'query' => "SELECT uid, first_name, pic_small FROM user WHERE uid IN (
    SELECT uid2 FROM friend WHERE uid1 = me() 
)")); 

// Loop through the rows 
// I think there is a page where you can test FQL at developers.facebook.com, I'll see if I can find it on my phone so I can see what the JSON response looks like and fix this code, in the mean while see if this works 

// Yeah so the code does this: 
// Loop through each item in the $allfriends result, (put the contents of the row into the $friend variable 
foreach($allfriends as $friend) 
{ 
    // This bit I need to change, this is supposed to look in the $likes array to see if the UID is present, but it will be comparing the uid against the whole object - need to tweak this 
    if(!in_array($friend['uid'], $likes)) 
    { 
     // Each friend that doesn't like the page should hit this part of the loop 
     echo $friend['uid']; 
    }     
} 

聲明 - 這是自從我使用PHP大約5年後，我已經觸摸了FQL的年齡 - 讓我知道是否有任何工作/不工作（我會很驚訝，如果它：D）