使用php

Question

我试图从数据库中获取一些信息并将其用作meta desc和keyword。但事情并没有像预期的那样发挥作用。

编辑：经过很多帮助，我得到了半工作。如果没有blogID，我希望它在另一部分上回落......脚本的那部分工作，任何想法？

<?php
$dsn = "sqlsrv:Server=localhost;Database=blog";
$conn = new PDO($dsn, "**********", "********");
$conn->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION );
$id = $_GET['postID'];
$sql = "SELECT * FROM blog_posts WHERE blogID=:id ORDER BY blogID DESC";
 $stmt = $conn->prepare($sql);
                    $stmt->execute (array($id));
                    while($metta = $stmt->fetch(PDO::FETCH_BOTH) )

if (isset($metta['blogID']) && !empty($metta['blogID'])) { 
$keywords = $metta['keywords']; 
$description = $metta['description']; 
} else { 
$keywords = "blalbalbalblabla"; 
$description = "blabla"; 
}

?>

Answer 1

它可能不是最漂亮的代码......但它有效！

<?php
$dsn = "sqlsrv:Server=localhost;Database=blog";
$conn = new PDO($dsn, "*****", "*******");
$conn->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION );
$id = $_GET['postID'];
$sql = "SELECT * FROM blog_posts WHERE blogID=:id";
 $stmt = $conn->prepare($sql);
                    $stmt->execute (array($id));
                    $metta = $stmt->fetch(PDO::FETCH_BOTH);

if (isset($_GET['postID'])) { 
$keywords = $metta['keywords']; 
$description = $metta['description']; 

}
sqlsrv_close($con);
?>


<meta name="description" content="<?= isset($description) ? $description : 'blablabla'; ?>">
        <meta name="keywords" content="<?= isset($keywords) ? $keywords : 'blablabla'; ?>">