PHP 带有 mysql 多项选择的准备语句没有 return 任何东西

Question

下面的 php 代码没有 return 任何东西，而下面的 return 值，唯一的区别是多选。这是为什么？（当我在我的浏览器中测试时，它没有拍摄任何东西）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();    
}

?>

一个选择的工作代码：

<?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

因为它 returns 多个值你需要绑定到多个变量。喜欢$stmt->bind_result($Col1, $Col2);

PHP 带有 mysql 多项选择的准备语句没有 return 任何东西

PHP prepared statement with mysql multiple selection doesn't return anything

php

mysql

prepared-statement