MySQL和PHP PDO如何獲取：SELECT EXISTS（從x WHERE y =：value中選擇SELECT 1）

Question

我使用這種語法而不是count（*），因為它應該更快，但我不知道如何獲取結果輸出

$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} 

但這似乎並沒有返回正確的東西，有什么想法嗎？

Answer 1

在這里使用EXISTS似乎是錯誤的。 只需執行以下查詢即可：

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

Answer 2

正確用法與通常一樣，捕獲fetch（）的返回值

$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
}