Question

我从我的代码中提取了这个片段，因为我担心有更好的方法可以做到这一点。我希望有人可以帮助我或指出我正确的方向。

基本上这个代码会检查数据库中的多个表，并检查结果是否返回。

    $stmt = $conn->prepare('SELECT email FROM 1table WHERE email = :email'); 
    $stmt = $conn->prepare('SELECT email FROM 2table WHERE email = :email');
    $stmt = $conn->prepare('SELECT email FROM 3table WHERE email = :email'); 
    $stmt->bindParam(':email', $email);
$stmt->execute();
    if($stmt->fetch(PDO::FETCH_NUM) > 0){

有更好的方法吗？或者更有效率的方式，可以告诉我结果在哪个表中找到了？

Answer 1

它是有效的PHP，但你的逻辑无效：

$stmt = $conn->prepare('SELECT email FROM 1table WHERE email = :email'); 
$stmt = $conn->prepare('SELECT email FROM 2table WHERE email = :email');
$stmt = $conn->prepare('SELECT email FROM 3table WHERE email = :email');

您继续使用新值覆盖$stmt。这意味着$stmt将只包含最后一个准备语句。

您可以查看UNION并修改您的代码，如下所示：

$sql = '(SELECT email FROM 1table WHERE email = :email)'; 
$sql .= 'UNION ALL'; 
$sql .= '(SELECT email FROM 2table WHERE email = :email)'; 
$sql .= 'UNION ALL'; 
$sql .= '(SELECT email FROM 3table WHERE email = :email)'; 
$stmt = $conn->prepare($sql);

Answer 2

这只会使用最后一个查询。你可以这样做：

$stmt = $conn->prepare('
    (SELECT email FROM 1table WHERE email = :email)
    union all
    (SELECT email FROM 2table WHERE email = :email)
    union all
    (SELECT email FROM 3table WHERE email = :email)');

Answer 3

以前的答案都没有告诉您如何知道哪个表来自哪个表。因此，如果这是相关的，那么这是正确的UNION

$sql  = "SELECT email, '1table' as fromTable  FROM 1table WHERE email = :email"; 
$sql .= " UNION ALL"; 
$sql .= " SELECT email, '2table' FROM 2table WHERE email = :email"; 
$sql .= " UNION ALL"; 
$sql .= " SELECT email, '3table' FROM 3table WHERE email = :email";

这个PDO代码有效吗？

3 个答案: