这个PDO代码有效吗? - Is this PDO code valid?

Is this PDO code valid?

本文关键字：有效代码 PDO 这个 | 更新日期: 2023-09-27

我已经从我的代码中提取了这个片段，因为我担心有更好的方法来做到这一点。我希望有人能帮助我或给我指出正确的方向。

基本上，这段代码检查数据库中的许多表，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?或者一种更有效的方式，它可以告诉我结果是在哪个表中找到的?

前面的答案都没有告诉您如何知道数据来自哪个表。所以，如果这是相关的，这就是正确的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";

这是有效的PHP，但你的逻辑是无效的:

$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');

继续用新值覆盖$stmt。这意味着$stmt将只包含最后一个prepare语句。

您可以查看UNION并按如下方式修改代码:

$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);

这将只使用最后一个查询。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)');