php说有些结果不是没有定义的,而有些是

polhcujo  于 2021-06-15  发布在  Mysql
关注(0)|答案(2)|浏览(289)

我有一个wierd php问题。我用下面的代码从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

但当我试图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之外的所有内容都可以打印出来;
注意:未定义的属性:stdclass::$yazar位于c:\users\nihal\documents\kutuphaneyonetim\php\body.php的第38行
注意:未定义的属性:stdclass::$yayinevi在c:\users\nihal\documents\kutuphaneyonetim\php\body.php的第39行
varèu dump($result)这样打印出来;

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" 
                } 
        }

有什么问题吗?你们能帮我找到吗?

bgibtngc

bgibtngc1#

你曾经

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

获取\u assoc,以便它们都作为数组返回。你把它们当作对象来处理,例如

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

alternativley将fetchall更改为

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

把代码保持原样。

klsxnrf1

klsxnrf12#

尝试将obj转换为array,这样如果您的值obj或值为null,它就不会出现任何错误,并且很容易管理array。
在循环之前添加。

<?php $result = json_decode(json_encode($result),true); ?>

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