If pagename then run statment

I’m building a blog and have two tables articles and categories and I’m joining these tables, but what I want to do is display just the categories if I click a categories link. I’m passing the categories_id in the link but I can’t figure out how to display just the categories if the page link is like blog.php?=category_id. I’ll have the categories_id number in the URL I just can’t figure out how to display just that category, I know the SQL statement to just display the categories but what I can’t figure out is how to run that statement if the URL contains the the?=category_id and not run my original SQL statement that was displaying all the articles by date. I tried to do if and else conditions depending on the page name but wasn’t working.

<?php
    connect_to_db();
    
    $url = $_SERVER['SCRIPT_NAME'];
    $pos = strrpos($url,"/");
    $pagename = substr($url,$pos+1);
    
    if($pagename == ("blog.php")) {
    $sql = "SELECT article_id, title, body, date, categories.category, author FROM articles
            LEFT JOIN categories ON articles.category = categories.category_id ORDER BY article_id DESC LIMIT 4";
    }
    elseif($pagename == ("blog.php?=1")) {
    $sql = "SELECT article_id, title, body, date, categories.category, author FROM articles
            LEFT JOIN categories ON articles.category = categories.category_id WHERE category_id = 1";
    }
    $result = query($sql);
    if($result===false) {
        echo "query failed";
    }
    else {
        while( $data = mysqli_fetch_array($result))
    {
?>     
    <article>
        <h3 class="title-medium"><?php echo $data['title']; ?></h3>
        <p class="caption-medium"><?php echo $data['author']; ?> <?php echo $data['date']; ?> <?php echo $data['category']; ?></p>
        <img src="img/blog-post-1.jpg" alt="">
        <p><?php echo substr($data['body'],0,450)." ..." ?></p>
        <a href="blog-post.php?id=<?php echo $data['article_id']; ?>"><p class="caption-medium highlight">Read More</p></a>
        <hr>
    </article>

    <?php
    }
}
?>

Answer

As you’ve been told in the comments, you have to use only parameter, not the whole request.

connect_to_db();

$where = '';
if (isset($_GET['cat_id'])) {
    $where = "WHERE category_id = ".intval($_GET['cat_id']);
}
$sql = "SELECT article_id, title, body, date, categories.category, author 
        FROM articles LEFT JOIN categories 
        ON articles.category = categories.category_id 
        $where ORDER BY article_id DESC LIMIT 4";

$result = query($sql);

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Answer