Question

我的div内容有些问题。我创建了一个网站，书籍必须从Mysql数据库加载到html div内容。我使用printf();来加载数据来分割内容。但我不能在<span>标签之间粘贴php变量。

    <?php
require 'scripts/connect.php';
$sql_select = "SELECT * FROM users";
$result = mysql_query($sql_select);
$row = mysql_fetch_array($result);
do
{
    printf(
                    '<ul>
                    <li>
                        <div class="product">
                            <a href="#" class="info">
                                <span class="holder">
                                    <img src="css/images/image01.jpg" alt="" />
                                    <span class="book-name">$row['b_name'] </span>     //ERROR IN HERE
                                    <span class="author">$row['a_name']</span>     //AND HERE
                                    <span class="description">Maecenas vehicula ante eu enim pharetra<br />scelerisque dignissim <br />sollicitudin nisi</span>
                                </span>
                            </a>
                            <a href="#" class="buy-btn">BUY NOW <span class="price"><span class="low">$</span>22<span class="high">00</span></span></a>
                        </div>
                    </li>
                    </ul>'
                    );
}
while($row = mysql_fetch_array($result));
         ?>

我怎么能这样做？感谢

Answer 1

如果你问我，我会这样做，没有mysql_*函数和分隔PHP和HTML 。您的问题是， '引号不会扩展变量。

<?php
require 'scripts/connect.php';
$sql_select = "SELECT * FROM users";
$result = mysqli_query($conn, $sql_select);
if (mysqli_num_rows($result) > 0)
    while (false != ($row = mysql_fetch_array($result))) { ?>
<ul>
    <li>
        <div class="product">
            <a href="#" class="info">
                <span class="holder">
                    <img src="css/images/image01.jpg" alt="" />
                    <span class="book-name"><?php echo $row['b_name']; ?></span>
                    <span class="author"><?php echo $row['a_name']; ?></span>
                    <span class="description">Maecenas vehicula ante eu enim pharetra<br />scelerisque dignissim <br />sollicitudin nisi</span>
                </span>
            </a>
            <a href="#" class="buy-btn">BUY NOW <span class="price"><span class="low">$</span>22<span class="high">00</span></span></a>
        </div>
    </li>
</ul>
<?php } ?>

在上面的代码中，我假设$conn是mysqli_connect变量。

Answer 2

问题在于您混合使用双引号和单引号。 PHP变量只能在双引号内输出。将所有现有的双引号更改为单引号，然后将<ul>封装在双引号中：

"<ul>
    <li>
      <div class='product'>
        <a href='#' class='info'>
          <span class='holder'>
            <img src='css/images/image01.jpg' alt='' />
            <span class='book-name'>$row['b_name']</span>
            <span class='author'>$row['a_name']</span>
            <span class='description'>Maecenas vehicula ante eu enim pharetra<br />scelerisque dignissim <br />sollicitudin nisi</span>
          </span>
        </a>
        <a href='#' class='buy-btn'>BUY NOW
          <span class='price'>
             <span class='low'>$</span>22
             <span class='high'>00</span>
          </span>
        </a>
      </div>
    </li>
 </ul>"

当然，您可以随时重构以简单地将内容输出为HTML，然后注入PHP变量：

<span><?php echo $row['b_name']; ?></span>

另请注意，{* 1}}调用在PHP 5.5中已弃用，而在PHP 7.0中已完全已删除，因此请考虑使用MySQLi或PDO：）

希望这有帮助！

如何在＆＃34; span＆＃34;中粘贴php变量标签？

2 个答案: