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我正在使用 PHP 会话变量创建一个购物车，除了购物车的显示之外，一切都很好。

我编写了以下代码以将购物车的内容显示为带有图像、名称、删除按钮和 id 的表格，问题是对于购物车中的每个项目，只有 id 不同，带有图像、名称和删除按钮都是指购物车中的第一项。

我猜我需要在 foreach 循环中包含查询，但不确定如何实现这一点，因为可能我只运行一次查询，但对 PHP 来说相对较新 - 非常感谢任何人的反馈 - 尝试学习当我走的时候！

<?php 
$array = explode(',', $_SESSION['cart']);
echo "<table border = '1'>";
foreach($array as $cartId) {
$ship_name = $row_ships['ship_name'];
$ship_image = $row_ships['image'];
echo "<tr>";
echo "<td><img src='images/ships/$ship_image'width='83' height='53' /> </td>";
echo "<td>$ship_name</td>";
echo "<td><a href='cart.php?action=delete&ship_id=$cartId'><img src='images/trash.png' width='31' height='42' /></a></td>";
echo "<td>$cartId</td>";
echo "</tr>";
} 
echo "</table>";
  ?>

这里也是查询，目前位于页面顶部

<?php
if (!function_exists("GetSQLValueString")) {
function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") 
{
  if (PHP_VERSION < 6) {
    $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue;
  }

  $theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue);

  switch ($theType) {
    case "text":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;    
    case "long":
    case "int":
      $theValue = ($theValue != "") ? intval($theValue) : "NULL";
      break;
    case "double":
      $theValue = ($theValue != "") ? doubleval($theValue) : "NULL";
      break;
    case "date":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;
    case "defined":
      $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
      break;
  }
  return $theValue;
}
}

$colname_ships = "-1";
if (isset($_SESSION['cart'])) {
  $colname_ships = $_SESSION['cart'];
}
mysql_select_db($database_ships, $ships);
$query_ships = sprintf("SELECT ship_id, ship_name, image FROM ship_infomation WHERE ship_id = %s", GetSQLValueString($colname_ships, "int"));
$ships = mysql_query($query_ships, $ships) or die(mysql_error());
$row_ships = mysql_fetch_assoc($ships);
$totalRows_ships = mysql_num_rows($ships);
?>
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1 回答 1

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根据您的代码，我假设 $_SESSION['cart'] = '1,2,3' （其中 1,2,3 是 ship_id）。

<?php
$query_ships = sprintf("SELECT ship_id, ship_name, image FROM ship_infomation WHERE ship_id IN (%s)", $_SESSION['cart']);
$ships = mysql_query($query_ships, $ships) or die(mysql_error());
$totalRows_ships = mysql_num_rows($ships);

echo "<table border = '1'>";
while($ship = mysql_fetch_array($ships)) {
  echo "<tr>";
  echo "<td><img src='images/ships/".$ship['image']."' width='83' height='53' /> </td>";
  echo "<td>".$ship['ship_name']."</td>";
  echo "<td><a href='cart.php?action=delete&ship_id=".$ship['ship_id']."'><img src='images/trash.png' width='31' height='42' /></a></td>";
  echo "<td>".$ship['ship_id']."</td>";
  echo "</tr>";
} 
echo "</table>";
?>
于 2013-07-28T09:23:00.020 回答