Question

以下代码工作正常，它从文件夹中随机选择一个图像，并将其指定为生成的Div的背景。这有效。如果我第二次使用这个代码，但是用它来简单地放置一个随机图像的img-src，这也有效。但是，以与第一个相同的方式使用代码会破坏网站。我做错了什么？

 <!-- BEGIN FEATURED 1 -->

                             <?php
try {
    $conn = new PDO("mysql:host=$dbhost;dbname=$dbname", $dbuser, $dbpass);
}
catch (PDOException $e) {
    echo $e->getMessage();
    echo 'Could not establish a connection to the database.';
}

$query = $conn->prepare('SELECT `articleid`,`title`  FROM `news_articles` WHERE  featured = 1 ORDER BY RAND() LIMIT 1');
$array = array(
    'N'
);
$query->execute();

$results = $query->fetchAll(PDO::FETCH_COLUMN, 0);
foreach ($results as $row) {
}

$image = get_rand_img('images/featured_images/csgo/');
$title = $result['title'];
echo '<a href="index.php?viewarticle=1&articleid=' . $row . '">';?>
<div id="featured-image" style="height: 267px; width: 292px; background:url(/images/featured_images/csgo/<?php echo $image ?>)">
<?php

$result = $conn->prepare("SELECT `articleid`,`title`,`short_title` FROM `news_articles` WHERE articleid=$row");
    $result->execute();
    $rows = $result->fetch();
        echo '<div class="featuredtitle1">';
        echo $rows['short_title'];
        echo '</div>';
        echo '</div>';
        echo '</a>';

?>

下面的代码打破了网站..（我现在已经评论过了，你可以看到）。

<!-- BEGIN FEATURED 2 (Disabled) -->

            <?php
/*
try {
    $conn = new PDO("mysql:host=$dbhost;dbname=$dbname", $dbuser, $dbpass);
}
catch (PDOException $e) {
    echo $e->getMessage();
    echo 'Could not establish a connection to the database.';
}

$query = $conn->prepare('SELECT `articleid` FROM `news_articles` WHERE  featured = 1 ORDER BY RAND() LIMIT 1');
$array = array(
    'N'
);
$query->execute();

$results = $query->fetchAll(PDO::FETCH_COLUMN, 0);
foreach ($results as $row) {
}

$image = get_rand_img('images/featured_images/dota/');
$title = $result['title'];
echo '<a href="index.php?viewarticle=1&articleid=' . $row . '">';?>
<div id="featured-image2" style="height: 267px; width: 292px; background:url(/images/featured_images/dota/<?php echo $image ?>)">
<?php

$result = $conn->prepare("SELECT `articleid`,`title`,`short_title` FROM `news_articles` WHERE articleid=$row");
    $result->execute();
    $rows = $result->fetch();
        echo '<div class="featuredtitle2">';
        echo $rows['short_title'];
        echo '</div>';
        echo '</div>';
        echo '</a>';
*/
?>

我不明白为什么第一段代码没问题，但重复它会再次破坏它？

Answer 1

在编辑评论之前，我看到了一个关于以数组形式访问对象的致命错误。尝试更改您的代码：

$results = $query->fetchAll(PDO::FETCH_COLUMN, 0);

对此：

$results = $query->fetchAll(PDO::FETCH_ASSOC);

两个几乎相同的代码片段......一个有效，另一个没有

1 个答案: