检查数据库上是否存在两个值，并通过一次咨询返回哪一个或两个值 - Check if two values exist on a database and return which one or both with one consultation

Check if two values exist on a database and return which one or both with one consultation

我正在尝试检查数据库中是否存在用户名或电子邮件，并将数据返回给用户，说明电子邮件或用户名或两者是否已在使用中，有什么方法可以在不对数据库进行两次咨询的情况下做到这一点？现在我有这个↓，但我不确定它是否正确。

 <?php
    if (isset($_POST['sub'])){
     require_once 'conexio.php';
     $pdo = new Conexio(); 
     $pdo->exec("set names utf8");
     $dbTabla="usuaris";
     $contra = $_POST["contra"];
     $consulta=$pdo->prepare("SELECT username,email FROM $dbTabla WHERE username=? OR email=?");
     $consulta->execute(array($_POST['username'],$_POST['email']));
     $result = $consulta->fetchObject();
     if($consulta->rowCount()>0){

     if("what to put in here?"){
      echo "That username it's already in use";
     }
     if("what to put in here?"){
      echo "That e-mail it's already in use";
     }
                             }

   }
   ?>

谢谢

您当前查询（以及 Chinnu 建议的解决方案）的"问题"是它可能返回最多两行（一行与用户名匹配，另一行与电子邮件匹配），您可能需要迭代，这可能有点烦人。

您可以使用 BIT_OR 聚合函数获得单行结果：

$consulta=$pdo->prepare("SELECT BIT_OR(username = ?) AS username_exists, " .
                        "BIT_OR(email = ?) AS email_exists " . 
                        "FROM $dbTabla " .
                        "WHERE username=? OR email=?");
$consulta->execute(array($_POST['username'], $_POST['email'],$_POST['username'], $_POST['email']));

现在，您有一行包含可以轻松计算的真/假表达式：

if($result->username_exists) {
    echo "That username is already in use";
}
if($result->email_exists) {
    echo "That e-mail is already in use";
}

你能试试这个吗，

    $result = $consulta->fetchObject();
    if($result->username==$_POST['username']){
            echo "That username it's already in use";
    }
    if($result->email==$_POST['email']){
        echo "That e-mail it's already in use";
    }