php - MySQL & php PDO 如何获取：SELECT EXISTS (SELECT 1 FROM x WHERE y = :value)

Question

我使用这种语法而不是 count (*) 因为它应该更快，但我不知道如何获取结果输出

$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} 

但它似乎没有返回正确的东西，知道吗？

Answer 1

这里的使用EXISTS似乎是错误的。只需执行此查询：

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

Answer 2

正确的用法和平常一样，捕获 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
}