MySQL＆php PDO如何獲取：SELECT EXISTS（SELECT 1 FROM x WHERE y =：value）

Question

我使用這種語法而不是count（*），因爲它應該更快，但我不知道如何獲取產生的輸出

$alreadyMember = $dataBase->prepare('SELECT EXISTS (SELECT 1 
FROM TheCommunityReachLinkingTable 
WHERE communityKey = :communityKey 
AND userID = :userID)'); 
$alreadyMember->bindParam(':communityKey', $_POST['communityKey'], PDO::PARAM_STR); 
$alreadyMember->bindParam(':userID', $_POST['userID'], PDO::PARAM_INT); 
$alreadyMember->execute(); 

if($alreadyMember->fetch()) {do code here} 

但它似乎沒有返回正確的東西，任何想法？

Answer 1

使用EXISTS在這裏似乎不對。只要執行這個查詢，而不是：

SELECT 1 
FROM TheCommunityReachLinkingTable 
WHERE communityKey = :communityKey 
AND userID = :userID 

Answer 2

正確的用法是一樣正常，捕捉取（）

$row = $alreadyMember->fetch(); 
print_r($row); // you may have numeric or string indices depending on the FETCH_MODE you set, pick one, consider a column alias and associative 


//or the easy way 
$alreadyMember->execute(); 
if ($alreadyMember-fetchColumn()) { 
    //column 0 contained a truthy value 
}