php – 在此脚本中使用PDO

2019年8月4日 192次阅读

我正在尝试使用PDO重写此脚本：

从数据库中获取值

mysql_select_db($database_db_FPF, $db_FPF);
$query_rsWorksMenu = "SELECT works.year, GROUP_CONCAT(workstitle.title) as titulos, GROUP_CONCAT(workstitle.id_title) as links FROM works JOIN workstitle ON works.id_year =      workstitle.id_year GROUP BY works.year ORDER BY works.year DESC";
$rsWorksMenu = mysql_query($query_rsWorksMenu, $db_FPF) or die(mysql_error());
$row_rsWorksMenu = mysql_fetch_assoc($rsWorksMenu);

显示值

     <ul>
        <?php do { ?>
        <li><span><?php echo $row_rsWorksMenu['year']; ?></span>
            <ul>
                <?php 
                 $titulos = explode(",", $row_rsWorksMenu['titulos']);
                 $links = explode(",", $row_rsWorksMenu['links']); 
                 foreach(array_combine($links, $titulos) as $link => $titulo){ 
                ?>
                <li><span><a href="works.php?id=<?php echo $link; ?>"><?php echo $titulo; ?></a></span></li>
                <?php } ?>
            </ul>
        </li>
        <?php }while ($row_rsWorksMenu = mysql_fetch_assoc($rsWorksMenu)); ?>
    </ul>

到目前为止我得到了什么：

从数据库中获取值：

$stmt2 = $conn->prepare('SELECT works.year, GROUP_CONCAT(workstitle.title) as titulos, GROUP_CONCAT(workstitle.id_title) as links FROM works JOIN workstitle ON works.id_year = workstitle.id_year GROUP BY works.year ORDER BY works.year DESC');
$stmt2->execute();
$result2 = $stmt2->setFetchMode(PDO::FETCH_OBJ);

显示值

        <ul>
        <?php do { ?>
        <li><span><?php echo $result2->year; ?></span>
            <ul>
                <?php 
                 $titulos = explode(",", $result2->titulos);
                 $links = explode(",", $result2->links); 
                 foreach(array_combine($links, $titulos) as $link => $titulo){ 
                ?>
                <li><span><a href="works.php?id=<?php echo $link; ?>"><?php echo $titulo; ?></a></span></li>
                <?php } ?>
            </ul>
        </li>
        <?php }while ($row_rsWorksMenu = mysql_fetch_assoc($rsWorksMenu)); ?>
    </ul>

问题是我不知道如何处理这一行：

}while ($row_rsWorksMenu = mysql_fetch_assoc($rsWorksMenu));

更新：

我改变了我的代码：

$stmt2 = $conn->prepare('...query...');
$stmt2->execute(); 
$result2 = $stmt2->fetchAll();

}while ($row_rsWorksMenu = $stmt2->fetchAll());

但没有任何反应.

更新2

我试试这段代码：

        <?php do { ?>
        <li><span><?php echo $result2->year; ?></span>
            <ul>
                <?php 
                 $titulos = explode(",", $result2->titulos);
                 $links = explode(",", $result2->links); 
                 foreach(array_combine($links, $titulos) as $link => $titulo){ 
                ?>
                <li><span><a href="works.php?id=<?php echo $link; ?>"><?php echo $titulo; ?></a></span></li>
                <?php } ?>
            </ul>
        </li>
        <?php }while($result2 = $stmt2->fetch()); ?>

我的DOM显示正确的< li>数.但没有价值观.为什么？

最佳答案 最简单的方法是这样的：

$sql = 'SELECT name, color, calories FROM fruit ORDER BY name';
foreach ($conn->query($sql) as $row) {
    print $row['name'] . "\t";
    print $row['color'] . "\t";
    print $row['calories'] . "\n";
}

PDO查询方法返回一个PDOStatement,您可以直接迭代.

http://www.php.net/manual/de/pdo.query.php

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