codeIgniter：将参数从先前的查询传递给选择查询

Question

我正在为浏览器游戏 travian 创建一个小管理工具。因此，我从数据库中选择所有村庄，并且想显示每个村庄特有的一些内容。但为了查询这些独特的详细信息，我需要传递村庄的 ID。我该怎么做？

这是我的代码（控制器）：

function members_area()
    {
        global $site_title;
        $this->load->model('membership_model');

        if($this->membership_model->get_villages())
        {
            $data['rows'] = $this->membership_model->get_villages();
            $id = 1;//this should be dynamic, but how?
            if($this->membership_model->get_tasks($id)):
                $data['tasks'] = $this->membership_model->get_tasks($id);
            endif;
        }

        $data['title'] = $site_title." | Your account";
        $data['main_content'] = 'account';
        $this->load->view('template', $data);
    }

这是我在模型中使用的 2 个函数：

function get_villages()
{
    $q = $this->db->get('villages');

    if($q->num_rows() > 0) {
        foreach ($q->result() as $row) {
            $data[] = $row;
        }
    return $data;
    }
}

function get_tasks($id)
{
    $this->db->select('name');
    $this->db->from('tasks');
    $this->db->where('villageid', $id);

    $q = $this->db->get();

    if($q->num_rows() > 0) {
        foreach ($q->result() as $task) {
            $data[] = $task;
        }
    return $data;
    }
}

当然还有视图：

<?php foreach($rows as $r) : ?>
        <div class="village">
            <h3><?php echo $r->name; ?></h3>
            <ul>
                <?php foreach($tasks as $task): ?>
                    <li><?php echo $task->name; ?></li>
                <?php endforeach; ?>
            </ul>
            <?php echo anchor('site/add_village/'.$r->id.'', '+ add new task'); ?>
        </div>
    <?php endforeach; ?>

ps：请不要删除第一段代码中的注释！

Answer 1

同意以上两点，加入吧:)

membership_model 功能：

function get_villages_and_tasks() {
  $query_results = $this->db
  ->select('villages.id,villages.name AS villagename,tasks.name AS taskname',false)
  ->join('tasks','tasks.villageid = villages.id','left')
  ->get('villages')->result_array();

  $return_array = array();

  foreach($query_results as $row) {
    if(!isset($return_array[$row['id']]) {
      $return_array[$row['id']] = array(
        'villagename'=>$row['villagename']
      );
    } 
    if(!empty($row['taskname'])) {
      $return_array[$row['id']]['tasks'][]=array('taskname'=>$row['taskname']);
    }
  }
  return $return_array;
}

查看：

<?php foreach($villages as $villageid=>$village) : ?>
  <div class="village">
    <h3><?php echo $village['villagename']; ?></h3>
    <ul>
      <?php if(!empty($village['tasks'])): ?>
        <?php foreach($village['tasks'] as $task): ?>
          <li><?php echo $task['taskname']; ?></li>
        <?php endforeach; ?>
      <?php endif; ?>
    </ul>
    <?php echo anchor('site/add_village/'.$villageid.'', '+ add new task'); ?>
  </div>
<?php endforeach; ?>

Answer 2

使用当前的模型函数，您可以迭代村庄，获取每个村庄的任务。

function members_area()
{
    global $site_title;
    $this->load->model('membership_model');

    // Fetch villages
    $villages = $this->membership_model->get_villages();

    if($villages)
    { 
        $tasks = array();

        // Iterate through villages fetching tasks
        foreach ($villages as $village)
        {
            $tasks[$village->id] = $this->membership_model->get_tasks($village->village->id);
        }

        $data['villages'] = $villages;
        $data['tasks'] = $tasks;
    }

    $data['title'] = $site_title." | Your account";
    $data['main_content'] = 'account';
    $this->load->view('template', $data);
}

但这需要 N+1 个 SQL 查询，其中 N 是村庄的数量。更好的解决方案是通过连接村表上的任务表来获取一个查询中的所有数据。