PHP 准备好的语句与 mysql 多重选择不返回任何内容

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下面的 php 代码不返回任何内容，而下面的代码返回值，唯一的区别是多重选择。这是为什么？(当我在浏览器中测试时，它没有拍摄任何内容)我也尝试将选择放在 () 之间，但没有帮助。

非工作代码:

<?php
$mysqli = new mysqli("x", "w", "y", "z");
$coresite = "Abbasya";
$rowx = "103";
$columnx = "3";
$directionx = "Back";
if($stmt = $mysqli->prepare("SELECT CABOwner, EtisaatTeam FROM CAB WHERE (SiteName=? AND Row=? AND Col=? AND Direction=?)"))
{
    $stmt->bind_param("ssss", $coresite, $rowx, $columnx, $directionx);
    $stmt->execute();
    $stmt->bind_result($cabinet);
    while ($stmt->fetch()) 
    {
        echo json_encode($cabinet).",";
    }
    $stmt->close();
}
else{
    $mysqli->close();    
}

？>

只需选择一项即可工作代码:

<?php
$mysqli = new mysqli("x", "w", "y", "z");
$coresite = "Abbasya";
$rowx = "103";
if($stmt = $mysqli->prepare("SELECT DISTINCT Col FROM CAB WHERE (SiteName=? AND Row=?)"))
{
    $stmt->bind_param("ss", $coresite, $rowx);
    $stmt->execute();
    $stmt->bind_result($Col);
    while ($stmt->fetch()) 
    {
        echo json_encode($Col).",";
    }
    $stmt->close();
}
else{
    $mysqli->close();    
}
?>

Answer 1

由于它返回多个值，因此您需要绑定(bind)到多个变量。就像$stmt->bind_result($Col1, $Col2);