带有 2 个表的 PHP PDO 更新

Question

我有一个小问题。这就是我想要实现的： 我有 2 个 mysql 表（类别、频道），频道表中有一个 cat_id。我想更新/编辑产品并将其放在另一个类别中，但我制作的代码仅显示一个类别 (id=1)，即使该产品的父 id(cat_id) 为 5。

try {
    //prepare query
    $query = "select channel_id, name, category_id from channels where channel_id = ? limit 0,1";
    $stmt = $pdo->prepare( $query );
    //this is the first question mark
    $stmt->bindParam(1, $_REQUEST['id']);
    //execute our query
    $stmt->execute();
    //store retrieved row to a variable
    $row = $stmt->fetch(PDO::FETCH_ASSOC);
    //values to fill up our form
    $channel_id = $row['channel_id'];
    $name = $row['name'];
    $category_id = $row['category_id'];
}catch(PDOException $exception){ //to handle error
    echo "Error: " . $exception->getMessage();
}
$query2 = "SELECT * FROM categories";
$stmt2 = $pdo->prepare( $query2 );
$stmt2->execute();
$results = $stmt2->fetchAll(PDO::FETCH_ASSOC);
?>
<!--we have our html form here where new user information will be entered-->
<form action='#' method='post' border='0'>
<table>
    <tr>
        <td>Channel Name</td>
        <td><input type='text' name='name' value='<?php echo $name;  ?>' /></td>
    </tr>
    <tr>
        <td>Category</td>
        <td>
        <?php foreach($results as $rows) {?>
        <select name="fileselect">
            <option name='cat_id' value=" <?php echo $rows['category_id']; ?>"> <?php echo $rows['name']; ?></option>
            <!-- <input type='text' name='category_id' value='<?php //echo $category_id;  ?>' /> -->
        <?php } ?>
        </select>
        </td>
    </tr>
    <tr>
        <td></td>
        <td>
            <!-- so that we could identify what record is to be updated -->
            <input type='hidden' name='channel_id' value='<?php echo $channel_id ?>' /> 
            <!-- we will set the action to edit -->
            <input type='hidden' name='action' value='update' />
            <input type='submit' value='Edit' />
        </td>
    </tr>
</table>
</form>

Answer 1

而不是

<?php foreach($results as $rows) {?>
    <select name="fileselect">
        <option name='cat_id' value=" <?php echo $rows['category_id']; ?>"> <?php echo $rows['name']; ?></option>
        <!-- <input type='text' name='category_id' value='<?php //echo $category_id;  ?>' /> -->
    <?php } ?>
    </select>

试试：

<select name="fileselect">
<?php foreach($results as $rows) {?>
    <option name='cat_id' value=" <?php echo $rows['category_id']; ?>"> <?php echo $rows['name']; ?></option>
    <!-- <input type='text' name='category_id' value='<?php //echo $category_id;  ?>' /> -->
<?php } ?>
</select>