如果我選擇該類別的新聞，如何選擇該類別的名稱

Question

我正在設置一個查詢，其中我選擇了新聞ID來顯示特定的ID新聞，我在新聞表中將類別表的外鍵作為類別ID，現在我希望如果顯示新聞，我還要顯示類別標題。

我試過拿起只顯示ID的類別ID，並且我想顯示該類別的標題。

<?php
        if(isset($_GET['news_id'])){
         $the_news_id = $_GET['news_id'];
        }
   $query = "SELECT * FROM news_title WHERE news_id = '$the_news_id'";
        $select_all_news_query = mysqli_query($connection,$query);
        while($row= mysqli_fetch_assoc($select_all_news_query)){
            $title = $row['news_title'];
            $description = $row['news_description'];
            $image = $row['news_image'];
            $news_cat_title= $row['news_cat_id'];
        ?>
        <h1 class="page-header">News Page</h1>

        <!-- First Blog Post -->
        <h2>
            <a href="#"><?php echo $title;?></a>
        </h2>


        <img class="img-responsive" src="image/<?php echo $image; ?>" 
alt="abc">
        <hr>
        <p><?php echo $description; ?></p>

        <hr>
<?php  }
        ?>
        <!-- Second Blog Post -->
        <hr>
        <div class="container">
            <div class="row">
                <h1 class="page-header">
  <!--This is where i want to show title but i am getting the ID--!>
                 <?php echo $news_cat_title; ?>
                </h1>
            </div>
        </div>

我希望得到標題，但我得到一個號碼。

Answer 1

在不知道各個表的架構的情況下，很難確定我是否正確解釋了該問題，但是我假設您要執行的操作是聯接兩個表？ 如果可以添加相關的表模式，則將有所幫助。

select * from news_title t
left outer join category c on c.category_id=t.category_id
where news_id = '$the_news_id'

根據您的最后評論，類別表簡稱為category ，外鍵為cat_id

select * from news_title t
left outer join category c on c.cat_id=t.cat_id
where news_id = '$the_news_id'

然后，當顯示記錄集時，您應該能夠：-

echo $row['cat_title'];

在Windows上捕獲和顯示表模式的簡單方法：

• 打開cmd提示
• 鍵入mysql -u root -p <DBNAME> 〜更改為實際數據庫！
• 類型describe <TABLE>; -按回車
• 復制顯示的信息並粘貼到文本編輯器中
• 對所有相關表重復

Answer 2

請嘗試以下代碼：

<?php
if(isset($_GET['news_id'])){
 $the_news_id = $_GET['news_id'];
}

$query = "SELECT N.*, NC.cat_id AS news_cat_id
FROM news_title AS N
LEFT JOIN news_category AS NC
ON N.cat_id = NC.cat_id 
WHERE news_id = '$the_news_id'";

$select_all_news_query = mysqli_query($connection,$query);

while($row= mysqli_fetch_assoc($select_all_news_query)){
    $title = $row['news_title'];
    $description = $row['news_description'];
    $image = $row['news_image'];
    $news_cat_title= $row['news_cat_id'];
?>
<h1 class="page-header">News Page</h1>

<!-- First Blog Post -->
<h2>
    <a href="#"><?php echo $title;?></a>
</h2>


<img class="img-responsive" src="image/<?php echo $image; ?>" alt="abc">
<hr>
<p><?php echo $description; ?></p>

<hr>
<?php } ?>
<!-- Second Blog Post -->
<hr>
<div class="container">
    <div class="row">
        <h1 class="page-header">
<!--This is where i want to show title but i am getting the ID--!>
         <?php echo $news_cat_title; ?>
        </h1>
    </div>
</div>

基於您的問題和評論的查詢語句可能像這樣：

$query = "SELECT N.*, NC.cat_id AS news_cat_id
FROM news_title AS N
LEFT JOIN news_category AS NC
ON N.cat_id = NC.cat_id 
WHERE news_id = '$the_news_id'";