请帮助：警告：mysqli_num_rows（）预计参数1被mysqli_result，给定对象

Question

我试着去创造这种动态表中的网站与MySQL

我都试过了，不能工作了什么是错，此代码。

请帮助

<?php 
// Script Error Reporting 
error_reporting(E_ALL); 
ini_set('display_errors', '1'); 
?> 
<?php 
// Run a select query to get my letest 6 items 
// Connect to the MySQL database 
include "storescripts/connect_to_mysql.php"; 
$dynamicList = ""; 
mysqli_query($con,"SELECT * FROM products ORDER BY date_added DESC LIMIT 6"); 
$productCount = mysqli_num_rows($con); // count the output amount 
if ($productCount > 0) { 
    while($row = mysqli_fetch_array($con)){ 
      $id = $row["id"]; 
      $product_name = $row["product_name"]; 
      $price = $row["price"]; 
      $date_added = strftime("%b %d, %Y", strtotime($row["date_added"])); 
      $dynamicList .= '<table width="100%" border="0" cellspacing="0" cellpadding="6"> 
     <tr> 
      <td width="17%" valign="top"><a href="product.php?id=' . $id . '"><img style="border:#666 1px solid;" src="inventory_images/' . $id . '.jpg" alt="' . $product_name . '" width="77" height="102" border="1" /></a></td> 
      <td width="83%" valign="top">' . $product_name . '<br /> 
      $' . $price . '<br /> 
      <a href="product.php?id=' . $id . '">View Product Details</a></td> 
     </tr> 
     </table>'; 
    } 
} else { 
    $dynamicList = "We have no products listed in our store yet"; 
} 
mysqli_close($con); 
?> 

Answer 1

您需要使用的mysqli_query()结果作为参数传递给mysqli_num_rows()，而不是连接。

$result = mysqli_query($con,"SELECT * FROM products ORDER BY date_added DESC LIMIT 6"); 
$productCount = mysqli_num_rows($result); 

这同样适用于mysqli_fetch_array()，它应该是

while($row = mysqli_fetch_array($result)){