php - 如何使用 PHP 错误显示 MySql 查询的多个结果

Question

我有一个商品类别查询，在本例中我有 2 种来自智利的 Wine 。我下面的代码工作正常，但只有 foreach 循环或无论它是什么，它会将第一个输出与第二个输出重叠。

我是 for-each 循环的新手

这是我的 PHP:

<?php  

include"db_connection.php";

$sql = mysql_query("SELECT * FROM WINE WHERE country='Chile'");

$allRows = array();
while($row = mysql_fetch_array($sql)) {
    $allRows[] = $row;
}       


foreach ($allRows as $row) {
            $id = $row ["id"];          
            $description = $row["description"];
            $wine_type = $row["wine_type"];
            $country = $row["country"];
            $bottle_price = $row["bottle_price"];
            $indicator = $row["indicator"];
            $colour = $row["colour"];
            $case_price = $row["case_price"];
            $case_size = $row["case_size"];      
            $date_added = strftime("%b %d, %Y", strtotime($row["date_added"]));

}


?>

这是 HTML:

<?php include('header.php'); ?>
<div id="content">
<table width="100%" border="0" cellspacing="0" cellpadding="15">
<?php
foreach ($allRows as $row) {
?>
    <tr>
      <td width="19%" valign="top"><img src="inventory_images/<?php echo $row['id']; ?>.jpg" width="142" height="188" alt="<?php echo $row['wine_type']; ?>" /><br />
      <a href="inventory_images/<?php echo $id; ?>.jpg">View Full Size Image</a></td>
            <td width="81%" valign="top"><h3><?php echo $wine_type; ?></h3>
            <p><?php echo "$".$bottle_price; ?><br /><br />
                <?php echo "$country $indicator"; ?> <br /><br />
                <?php echo $description; ?> <br />
            </p>

              <form id="form1" name="form1" method="post" action="cart.php">
                <input type="hidden" name="pid" id="pid" value="<?php echo $id; ?>" />
                <input type="submit" name="button" id="button" value="Add to Shopping Cart" />
              </form>
            </td>
    </tr>
<?php
}
?>
</table>
</div>
<?php include('footer.php'); ?>

Answer 1

发生的情况是您覆盖了您的 $id等每个循环上的变量，因此只有最后一行的结果被保存在那里。

但是，您根本不需要这些变量，因为您已将所有信息保存在 $allRows 中。 。就像你通过<?php echo $row['id']; ?>访问id一样在 HTML 中，您应该处理所有其他变量。例如:$row["bottle_price"]而不是$bottle_price .

您的 PHP 代码将如下所示:

<?php 
    include"db_connection.php";

    $sql = mysql_query("SELECT * FROM WINE WHERE country='Chile'");
    $allRows = array();

    while($row = mysql_fetch_array($sql)) {
        $allRows[] = $row;
    }
?>

你的 HTML 像这样:

<?php include('header.php'); ?>
<div id="content">
<table width="100%" border="0" cellspacing="0" cellpadding="15">
<?php
    foreach ($allRows as $row) {
?>
    <tr>
        <td width="19%" valign="top">
            <img src="inventory_images/<?php echo $row['id']; ?>.jpg" width="142" height="188" alt="<?php echo $row['wine_type']; ?>" /><br />
            <a href="inventory_images/<?php echo $row['id']; ?>.jpg">View Full Size Image</a>
        </td>
        <td width="81%" valign="top">
            <h3><?php echo $row['wine_type']; ?></h3>
            <p><?php echo "$".$row['bottle_price']; ?><br /><br />
                <?php echo $row['country']." ".$row['indicator']; ?> <br /><br />
                <?php echo $row['description']']; ?> <br />
            </p>

            <form id="form1" name="form1" method="post" action="cart.php">
                <input type="hidden" name="pid" id="pid" value="<?php echo $row['id']; ?>" />
                <input type="submit" name="button" id="button" value="Add to Shopping Cart" />
            </form>
        </td>
    </tr>
<?php
    }
?>
</table>
</div>
<?php include('footer.php'); ?>