Question

试图回显2个变量，但是两个值都得到了“意外的索引”。我认为回声有问题。我在错误地要求$ faktura对象中的值。但是我不知道我在做什么错。

这是我的错误消息：

Notice: Undefined index: userID in C:\Faktura\scripts\faktura.php on line 21

Notice: Undefined index: fakturaNr in C:\Faktura\scripts\faktura.php on line 22


faktura.php
<?php 
session_start();
require("config/database.php");

if(isset($_POST['fakturaBtn'])){

    $fakturaNr = 50163;
    $userID = 144;

    $stmt = $pdo->prepare('SELECT * FROM myView WHERE fakturaNr = :fakturaNr AND userId = :userID');
    $stmt->execute(['fakturaNr' =>  $fakturaNr, 'userID' => $userID]); //Where the exception occurs
    $faktura = $stmt->fetch();

    echo $faktura['userID'];
    echo $faktura['fakturaNr'];
}
?>

database.php
<?php
$host = 'myhost';
$db   = 'mydb';
$user = 'myuser';
$pass = 'mypass';

$dsn = "sqlsrv:Server=$host;Database=$db";
//$dsn = "sqlserv:host=$host;dbname=$db;charset=$charset";
$options = [
    PDO::ATTR_ERRMODE              => PDO::ERRMODE_EXCEPTION,
    PDO::ATTR_DEFAULT_FETCH_MODE   => PDO::FETCH_ASSOC,
    PDO::SQLSRV_ATTR_ENCODING      => PDO::SQLSRV_ENCODING_UTF8,
];
try {
     $pdo = new PDO($dsn, $user, $pass, $options);
} catch (\PDOException $e) {
     throw new \PDOException($e->getMessage(), (int)$e->getCode());
}
?>

我尝试使用带有isset的if语句，但这也不起作用。这是我尝试的方法

if(isset($faktura)) {
    echo $faktura['userID'];
    echo $faktura['fakturaNr'];
}

未定义索引：用于数据库中的变量

0 个答案: