Question

这里我无法从数据库中获取数据。我想要结果，因为当我点击类别时，结果应来自产品数据库，其中c_id =来自数据库的类别这里显示的菜单来自我的数据库表名=类别当点击时，我应该将结果作为硬币选择器=得到页面右侧（与产品表中的硬币选择器相关的结果）。我的类别数据库表是产品表的数据库表是

<?php    
    $link = mysqli_connect("localhost", "root", "", "admin_login");
     include 'header.php';
    ?>  
     <div class="container-fluid">
         <div class="row">
             <div class="col-lg-offset-1 col-lg-2">
                 <div class='panel panel-default'>
                     <?php 
                        $query = "SELECT * FROM category";  
                        $res = mysqli_query($link, $query);  
                        while($row = mysqli_fetch_array($res))  
                        {  
                             ?>
                         <div class='nav nav pills nav-stacked nav_custom'>
                         <li><a href='category.php?id=<?php echo $row['c_id']; ?>'><?php echo $row['c_name']; ?></a></li>
                     </div>
                             <?php
                        }  
                     ?>
                 </div>
             </div>
             <div class='col-lg-6'>
                 <div class="panel panel-info">
                                <div class="panel panel-body">
                                    <?php
                                    $c_id = $row["c_id"];
                                    $query = "SELECT * FROM product WHERE c_id='.$c_id.' ";  
                                    $res = mysqli_query($link, $query);  
          while($row = mysqli_fetch_array($res))  
          {  ?>
              <div class="col-lg-3">  
              <div class="panel panel-info">  
              <div class="panel-heading"><?php $row["product_name"]; ?> 
              </div>
              <div class="panel-body"> 
          <center> 
              <img src="image/product_image/'<?php echo $row["product_img"]; ?>'" alt="'<?php $row["product_name"]; ?> '" class="img-responsive" width="200px" height="200px">
              </center> 
          </div> 
              <div class="panel-footer">           
              <a href="product_details.php?id=<?php echo $row["id"]; ?>" class="btn btn-info btn-block btn-xs">Details           
              </a>
              <a href="contact.php?id=<?php $row["id"]; ?> " class="btn btn-info btn-block btn-xs"><span class="glyphicon glyphicon-credit-card"></span>&nbsp;Request Sample Quote           
              </a>
              </div>  
              </div>  
              </div> 
          <?php } ?>
                                </div>
                            </div>
                        </div>
             </div>
         </div>
     </div>
      <p><br></p>
     <p><br></p>
     <p><br></p>
    <?php
    include 'footer.php';
    ?>

Answer 1

不是它的可变覆盖问题

你的查询中有语法错误。没有它不会覆盖问题。你传递变量id以显示产品的问题看到这里

<li><a href='category.php?id=<?php echo $row['c_id']; ?>'>

并且您正在尝试使用$c_id获取不存在的产品。请参阅此处

<div class="panel panel-body">
<?php
$c_id = $row["c_id"];
query = "SELECT * FROM product WHERE c_id='.$c_id.' ";  
$res = mysqli_query($link, $query);

然后你需要像这样访问id

    if (isset($_GET['id'])) {
        $c_id = $_GET['id'];
       $query = "SELECT * FROM product WHERE c_id='$c_id' ";  
       $res = mysqli_query($link, $query); 
      //your `while` loop 
    }

请记住，您的查询对sql注入攻击使用prepared statement

是开放的

完整代码

<?php    
$link = mysqli_connect("localhost", "root", "", "admin_login");
include 'header.php';
?>  
<div class="container-fluid">
<div class="row">
<div class="col-lg-offset-1 col-lg-2">
<div class='panel panel-default'>
<?php 
$query = "SELECT * FROM category";  
$res = mysqli_query($link, $query);  
while($row = mysqli_fetch_array($res)){ ?>
    <div class='nav nav pills nav-stacked nav_custom'>
    <li><a href="category.php?id=<?php echo $row['c_id']; ?>">
    <?php echo $row['c_name']; ?></a></li>
    </div>
    <?php
}  
?>
</div>
</div>
<div class='col-lg-6'>
<div class="panel panel-info">
<div class="panel panel-body">
<?php
if (isset($_GET['id'])) {
    $c_id = $_GET['id'];
    $query = "SELECT * FROM product WHERE c_id='$c_id' ";  
    $res = mysqli_query($link, $query);  
    while($row = mysqli_fetch_array($res)) {  ?>
        <div class="col-lg-3">  
        <div class="panel panel-info">  
        <div class="panel-heading"><?php $row["product_name"]; ?> 
        </div>
        <div class="panel-body"> 
        <center> 
        <img src="image/product_image/'<?php echo $row["product_img"]; ?>'" alt="'<?php $row["product_name"]; ?> '" class="img-responsive" width="200px" height="200px">
        </center> 
        </div> 
        <div class="panel-footer">           
        <a href="product_details.php?id=<?php echo $row["id"]; ?>" class="btn btn-info btn-block btn-xs">Details           
        </a>
        <a href="contact.php?id=<?php $row["id"]; ?> " class="btn btn-info btn-block btn-xs"><span class="glyphicon glyphicon-credit-card"></span>&nbsp;Request Sample Quote           
        </a>
        </div>  
        </div>  
        </div> 
    <?php } 
} ?> 
</div>
</div>
</div>
</div>
</div>
</div>
<p><br></p>
<p><br></p>
<p><br></p>
<?php
include 'footer.php';
?>

单击动态数据库后，不显示结果

1 个答案: