Question

我正在尝试创建一个列出MySQL数据库中最近6个项目的页面。我使用以下代码（gen列是包含图像URL的列）;

        $result = mysqli_query($con,"SELECT gen,title,view FROM id
        ORDER BY 'id' DESC LIMIT 5,1");

        $row = mysql_fetch_array($result)

            while ($row = mysql_fetch_array($result)){
            echo '
            <div class="row">
                <div class="col-md-4 col-sm-6">
                    <div class="portfolio-item">
                        <div class="portfolio-image">
                            <a href="page-portfolio-item.html"><img src="i/raw' . $row['gen'] . ' alt="' . $row['title'] . '"></a>
                        </div>
                        <div class="portfolio-info-fade">
                            <ul>
                                <li class="portfolio-project-name">' . $row['title'] . '</li>
                                <li>Views: ' . $row['view'] . '</li>
                                <li class="read-more"><a href="http://minecraftimg.com/i/' . $row['gen'] . '.php" class="btn">Go!</a></li>
                            </ul>
                        </div>
                    </div>
                </div>';
                }

但是，我收到以下错误;

PHP Parse error:  syntax error, unexpected 'while' (T_WHILE) in /home/...

任何帮助将不胜感激！

Answer 1

缺少一个;在你的while循环之前：

$row = mysql_fetch_array($result);

Answer 2

缺少分号：

$row = mysql_fetch_array($result);

解析错误，同时声明（PHP）

2 个答案: