类别和子类别不匹配

Question

我对类别和子类别字段有疑问。

不久，我做了这样的分类：文学，并想制作这样的下拉子菜单：诗歌，小说，非小说。

我在网站上收到所有类别，但根本无法获得子类别字段。

这是我的代码：

     $category=htmlspecialchars($_REQUEST['category']);
if(isset($category)){
    $avt=mysql_query("select * from  category order by id asc");
    while($category_id=mysql_fetch_array($avt))
    {
        ?> 

        <li><a href="#"> <k style="font-size:11px;"> <? 
   echo $category_id['catname'];            
    } }
    ?>  </k> </a> 




     <?



     $subcategory=htmlspecialchars($_REQUEST['subcategory']);
if(isset($subcategory)){
    $sql = mysql_query("SELECT t1.*,t2.* FROM category as t1, subcategory as t2 WHERE t1.id=t2.'$_REQUEST[subcat_id]'");
while($data = mysql_fetch_array($sql))

    { ?>


    <ul style="z-index:101;">
   <li> <k>  <a href="index.php?do=cat&id=<? echo $data['id'];?>" style="border-bottom:1px solid #CECECE; position:relative; left:-40px; font-size:11px; ">&nbsp; <?php echo $data["subcat_name"]; ?> </a> </k></li> 

   <?  } }

    ?>

这是mysql：

类别

id
catname

子类别

id
subcat_id
subcat_name

就这些。如果有人可以帮助我解决这个问题，我将不胜感激。 （显然我是初学者）

谢谢

Answer 1

即使您是初学者，我建议您也可能会对此进行大量调整。从数据库连接开始。 mysql_函数在最新的PHP v5中已弃用（不建议使用），在PHP v7中将其完全删除，因为它们存在安全风险，并且大多数人都无法很好地实现。请改用PDO或MySQLi。我更喜欢PDO，因此我将演示：

# Create the connection thusly, fill in your own credentials
$con = new \PDO("mysql:host=localhost;dbname=mydatabase", $username, $password);

接下来，您应该知道isset和empty之间的区别。在您的情况下，如果未设置isset，则$_REQUEST['category']将发出警告。当您创建自己的变量时，保证可以在检查它时将其设置，因为您是手动创建的。

如果未设置$_REQUEST['category']，则将发出警告，因为在检查之前已分配了它：

$category = htmlspecialchars($_REQUEST['category']);

此操作不会发出警告，因为您正在检查是否已设置$_REQUEST['category']。如果不是，请默认将$category分配给false：

# I like to use a ternary when there are only two options, but you can use if/else
$category = (isset($_REQUEST['category']))? htmlspecialchars(trim($_REQUEST['category'])) : false;

# This part now checks if the value that you created is NOT EMPTY because you
# have just created it, so checking if it is set is illogical
if(!empty($category)):
    # Now we use our connection here to query.
    # Here we can use query() because we don't have any variables to add into the query,
    # so it's safe to use this short version
    $avt = $con->query("select * from  category order by id asc");
    # Same while loop, just use the PDO version of fetching the results
    while($category_id = $avt->fetch(\PDO::FETCH_ASSOC)): ?>

        <?php
        /**
         * I'm not sure why you are using a "k" tag, so just combine with "a" tag
         * also, though you can use <? as a php open, it's not recommended
         * lastly, you have the close tag for "a" outside the loop, that is bad syntax
         * since the open is inside the loop
         */
        ?>

        <li><a href="#" style="font-size: 11px;"><?php echo $category_id['catname'] ?></a></li>

    <?php endwhile ?>
<?php endif ?>

现在顶部完成了，在代码的第二部分上您遇到了相同的问题（还有更多的问题：

# It's also worth noting here you have set this variable, but you don't actually use it...?
# Secondly, you are checking for "subcategory" here, but in the code below you are injecting "subcat_id"...should they match??
$subcategory = (isset($_REQUEST['subcategory']))? htmlspecialchars(trim($_REQUEST['subcategory'])) : false;

if(!empty($subcategory)):
    # Here is where you need to make a detour with your script. What you are doing is unsafe.
    # You can not inject that value right into the sql. You need to "prepare", "bind value", and "execute" it
    $subcat_id = $_REQUEST['subcat_id'];
    # First prepare using ? as placeholders for the value you want to search for
    # The naming convention on the "subcat_id" column I assume is actually the
    # the parent category? If so, maybe it should be called "parent_id" or "category_id"??
    $sql = $con->prepare("SELECT t1.*, t2.* FROM category as t1, subcategory as t2 WHERE t1.id = ? AND t2.subcat_id = ?");
    # Now you want to execute for those two question marks (?)
    # There are a few ways to prepare and bind parameters, this way is easy
    # and you can visually see what we are trying to accomplish
    $sql->execute([$subcat_id, $subcat_id]) ?>

    <!-- you probably don't want to be looping this (or it's close tag) -->
    <ul style="z-index:101;">

    <?php
    # Same type of thing as your first block of code
    while($data = $sql->fetch(\PDO::FETCH_ASSOC)): ?>

        <li><a href="index.php?do=cat&id=<?php echo $data['id'] ?>" style="border-bottom:1px solid #CECECE; position:relative; left:-40px; font-size:11px; ">&nbsp;<?php echo $data["subcat_name"] ?> </a></li> 

    <?php endwhile ?>

    <!-- end list -->
    </ul>
<?php endif ?>

这里的最终结论是，在遇到主要问题（可能是您的SQL语句和所查询的值）之前，您需要做一些工作。我建议您阅读您不熟悉的任何提及的概念。