Php 如何使用CodeIgniter在单个post中连接表和显示

我有三张桌子

职位

• 身份证
• 猫咪
• 用户id
• 头衔
• 内容

使用者

• 身份证
• 用户名
• 通过

类别

• 身份证
• 名字

我有一个代码如下：

我的帖子管理员

$data['all_categories'] = $this->category_model->get_all_categories();
$data['posts'] = $this->post_model->get_posts();
$data['users'] = $this->user_model->get_user();

$this->load->view('pages/template', $data);

$data['posts'] = $this->post_model->load_posts();

我的post\u模型

$this->db->order_by('posts.id', 'DESC');
$this->db->join('categories', 'categories.id = posts.category_id');
$this->db->join('users', 'users.id = posts.user_id');
$query = $this->db->get('post');
return $query->result_array();

public function load_posts(){
    $this->db->select('tbl_p.*, tbl_u.username, tbl_u.pass, tbl_c.name as category_name');        
    $this->db->from('posts as tbl_p');
    $this->db->join('users as tbl_u', 'tbl_u.id = tbl_p.user_id', 'left');
    $this->db->join('categories as tbl_c', 'tbl_c.id = tbl_p.cat_id', 'left');
    $this->db->order_by('tbl_p.id', 'DESC');        
    $query = $this->db->get();
    return ($query->num_rows() > 0) ? $query->result_array() : false;        
}

我的索引视图

<?php foreach ($posts as $post): ?>
    <br>
    <div class="media">
        <div class="media-body" style="margin-left: 5px;">
            <h5><?php echo $post['title']; ?></h5>
            <small class="post-date"><span class="fa fa-user"></span> <strong><?php echo $post['username']; ?></strong> | <span class="fa fa-tags"></span> <strong><?php echo $post['name']; ?></strong></small><br>
            <?php echo character_limiter($post['content'], 250); ?>
            <hr>
        </div>
    </div>
<?php endforeach; ?>

<h5><?php echo $post['title']; ?></h5>
<small class="post-date"><span class="fa fa-user"></span> <strong><?php echo $post['username']; ?></strong> | <span class="fa fa-tags"></span> <strong><?php echo $post['category_id']; ?></strong></small><br>
<?php echo $post['content']; ?>

|

我成功地在索引视图中显示了它，但用户名和类别名称在单个帖子中不起作用

我的帖子视图

<?php foreach ($posts as $post): ?>
    <br>
    <div class="media">
        <div class="media-body" style="margin-left: 5px;">
            <h5><?php echo $post['title']; ?></h5>
            <small class="post-date"><span class="fa fa-user"></span> <strong><?php echo $post['username']; ?></strong> | <span class="fa fa-tags"></span> <strong><?php echo $post['name']; ?></strong></small><br>
            <?php echo character_limiter($post['content'], 250); ?>
            <hr>
        </div>
    </div>
<?php endforeach; ?>

<h5><?php echo $post['title']; ?></h5>
<small class="post-date"><span class="fa fa-user"></span> <strong><?php echo $post['username']; ?></strong> | <span class="fa fa-tags"></span> <strong><?php echo $post['category_id']; ?></strong></small><br>
<?php echo $post['content']; ?>

|

---

请帮帮我，伙计们。谢谢

在文件中添加以下代码

在型号中

$this->db->order_by('posts.id', 'DESC');
$this->db->join('categories', 'categories.id = posts.category_id');
$this->db->join('users', 'users.id = posts.user_id');
$query = $this->db->get('post');
return $query->result_array();

public function load_posts(){
    $this->db->select('tbl_p.*, tbl_u.username, tbl_u.pass, tbl_c.name as category_name');        
    $this->db->from('posts as tbl_p');
    $this->db->join('users as tbl_u', 'tbl_u.id = tbl_p.user_id', 'left');
    $this->db->join('categories as tbl_c', 'tbl_c.id = tbl_p.cat_id', 'left');
    $this->db->order_by('tbl_p.id', 'DESC');        
    $query = $this->db->get();
    return ($query->num_rows() > 0) ? $query->result_array() : false;        
}

在控制器中

$data['all_categories'] = $this->category_model->get_all_categories();
$data['posts'] = $this->post_model->get_posts();
$data['users'] = $this->user_model->get_user();

$this->load->view('pages/template', $data);

$data['posts'] = $this->post_model->load_posts();

---

可能是因为行名称相同。对于您选择的每一行，使用SELECT作为“唯一名称”。

post.category\U id应为post.category\U id（在post\U模型中）我已将其替换，但结果保持不变。