php 说有些结果不是未定义的，有些是

Question

我有一个奇怪的 PHP 问题。我使用下面的代码从 mysql 获取数据

SELECT kitap.kit_id, kitap.kit_adi, yazar.yad as yazar,  
        yayinevi.yeviad as yayinevi,kitap.isbn,kitap.basim_yili, 
        kitap.baski_no,kitap.dil,kitap.cilt,kitap.sayfa, 
        kitap.kategori,kitap.durum
FROM kitap 
    INNER JOIN kitapyayinevi ON kitap.kit_id = kitapyayinevi.kit_id 
    INNER JOIN kitapyazar ON kitap.kit_id = kitapyazar.kit_id 
    INNER JOIN yayinevi ON kitapyayinevi.yevi_id = yayinevi.yevi_id 
    INNER JOIN yazar ON kitapyazar.yazar_id = yazar.yazar_id
SQL;

    $stmt = $connection->query($sql);
    $result = $stmt->fetchAll(PDO::FETCH_ASSOC);
    var_dump($result); //This is for debugging purposes

但是当我尝试用下面的代码打印它们时；

<?php foreach ($result as $item): ?>
    <tr>
        <td class="hide"><?= $item->kit_id; ?></td>
        <td><?= $item->kit_adi; ?></td>
        <td><?= $item->yazar; ?></td>
        <td><?= $item->yayinevi; ?></td>
        <td><?= $item->isbn; ?></td>
        <td><?= $item->baski_no; ?></td>
        <td><?= $item->basim_yili; ?></td>
        <td><?= $item->dil; ?></td>
        <td><?= $item->cilt; ?></td>
        <td><?= $item->sayfa; ?></td>
        <td><?= $item->kategori; ?></td>

它给了我下面的错误。除了 $item->yazar 和 $item->yayinevi 之外的所有内容都打印出来了；

Notice: Undefined property: stdClass::$yazar in C:\Users\Nihal\Documents\KutuphaneYonetim\php\body.php on line 38

Notice: Undefined property: stdClass::$yayinevi in C:\Users\Nihal\Documents\KutuphaneYonetim\php\body.php on line 39

然后 var_dump($result) 打印出来是这样的；

array(1) { [0]=> array(12) { 
                ["kit_id"]=> string(2) "25" 
                ["kit_adi"]=> string(7) "Sınır" 
                ["yazar"]=> string(13) "Beyza Alkoç " 
                ["yayinevi"]=> string(15) "İNDİGO KİTAP" 
                ["isbn"]=> string(10) "6052361580" 
                ["basim_yili"]=> string(4) "2018" 
                ["baski_no"]=> string(1) "4" 
                ["dil"]=> string(8) "TÜRKÇE" 
                ["cilt"]=> string(12) "Karton Kapak" 
                ["sayfa"]=> string(3) "400" 
                ["kategori"]=> string(5) "Roman" 
                ["durum"]=> string(5) "rafta" 
                } 
        }

会出现什么问题？你们能帮我找到吗？

Answer 1

你用过

$result = $stmt->fetchAll(PDO::FETCH_ASSOC);

FETCH_ASSOC 所以它们都作为数组返回。您将它们当作对象来处理，例如使用

<td><?= $item['yazar']; ?></td>

或者将 fetchAll 更改为

$result = $stmt->fetchAll(PDO::FETCH_OBJ);

并保留原来的代码。