Question

其他一切都应该没问题，但是我仍然在显示else功能时遇到问题。理论上我相信一切都应该有效。我查看了所有语法，一切看起来都很好，一切都已正确关闭。我想要做的是，如果$_GET['cat'] == 1, 2, or 3（这就是为什么我把它作为一个数组）是有效的，那么我希望能够拉出那个特定的类别，如果没有那么我希望它显示所有的类别在一起。很像WordPress的功能。

<?php

include('config.inc.php');
// security check to make sure it's not grabbing anything extra that's not there
$get_cat_id = mysqli_query($mysql_conn, "SELECT `id` FROM `categories`");
$cat_array = array();
while($row = mysql_fetch_assoc($get_cat_id)) {
    $cat_array[] = $row['id'];
}
echo $cat_array[];

// checks to see if it's calling for a specific category
if(isset($_GET['cat']) && ($_GET['cat'] == $cat_array)) {
    $fetch_post_content = mysqli_query($mysql_conn, "SELECT * FROM `posts` WHERE `category_id` = '".$_GET['cat']."' ORDER BY `id` DESC");
    $fetch_category_content = mysqli_query($mysql_conn, "SELECT * FROM `categories` WHERE `id` = '".$_GET['cat']."'");
?>
        <h1 class="category-title"><?= $fetch_category_content->name; ?></h1>
<?php
    while($post_content = mysqli_fetch_object($fetch_post_content)) { ?>
        <article>
            <?php
            // this checks for and grabs the featured image from the database
            if(isset($post_content->img_id)) {
                $fetch_post_image = mysqli_query($mysql_conn, "SELECT * FROM `images` WHERE `id` = '".$post_content->img_id."'");
            ?>
            <img src="<?= $fetch_post_image->url; ?>">
            <?php } ?>

            <h3 class="post-title"><?= $post_content->title; ?></h3>
            <h5 class="post-datetime">Posted on <?= $post_content->date; ?> at <?= $post_content->time; ?></h5>
            <p class="post-content"><?= $post_content->content; ?></p>
            <?php
            // grabs authors name from user database
            $fetch_post_author = mysqli_query($mysql_conn, "SELECT * FROM `users` WHERE `id` = '".$post_content->author_id."'");
            ?>
            <h6 class="post-author">Posted by <?= $fetch_post_author->name; ?></h6>


        </article>

<?php
    // ends while function
    }

// displays all conent for the front page if no specific category is called
} else {
    $fetch_post_content = mysqli_query("SELECT * FROM `posts` ORDER BY `id` DESC");
    while($post_content = mysqli_fetch_object($fetch_post_content)) {
        ?>

        <article>
            <?php
            // this checks for and grabs the featured image from the database
            if(isset($post_content->img_id)) {
                $fetch_post_image = mysqli_query($mysql_conn, "SELECT * FROM `images` WHERE `id` = '".$post_content->img_id."'");
            ?>
            <img src="<?= $fetch_post_image->url; ?>">
            <?php } ?>

            <h3 class="post-title"><?= $post_content->title; ?></h3>
            <h5 class="post-datetime">Posted on <?= $post_content->date; ?> at <?= $post_content->time; ?></h5>
            <p class="post-content"><?= $post_content->content; ?></p>
            <?php
            // grabs authors name from user database
            $fetch_post_author = mysqli_query($mysql_conn, "SELECT * FROM `users` WHERE `id` = '".$post_content->author_id."'");
            ?>
            <h6 class="post-author">Posted by <?= $fetch_post_author->name; ?></h6>


        </article>
        <?php

    }
}


?>

我错过了什么吗？

Answer 1

您无法将标量变量与数组进行比较。您需要我们in_array()：

if(isset($_GET['cat']) && in_array($_GET['cat'], $cat_array)) {

正如Fred指出的那样，在你修复mysql和mysqli函数的混合之前，它无论如何都不会起作用。

我的if语句无法正常工作的一些原因

1 个答案: