Php CodeIgniter将带有查询的数组传递到视图中

您好，我在获取视图以显示以下代码的任何结果时遇到问题

控制器：

   $datedue = 2011-01-27;
  $username = $this->tank_auth->get_username();
  $sql = "SELECT taskid FROM relationships WHERE username = ? AND datedue = ?"; 
  $tasks = $this->db->query($sql, array($username, $datedue));
  $count = 0;
      $taskdetails = array();

  foreach($tasks->result() as $row)
  {

   $taskid = $row->taskid;

    $subsql = "SELECT * FROM tasks WHERE id = ?"; 
    $taskdetails[$count] = $this->db->query($subsql, array($taskid));
    $count++;
  }

  $data['taskdetails'] = $taskdetails;
  $data['total'] = $count;

  $this->header();
  $this->load->view('dashboard', $data);

    <?php 

 foreach($taskdetails as $entry)
   {
  foreach($entry->result() as $row)
  {
   echo $row->name;
  }

   }

?>

   foreach($taskdetails as $task)
   {
     echo $task->name;
   }

查看：

   $datedue = 2011-01-27;
  $username = $this->tank_auth->get_username();
  $sql = "SELECT taskid FROM relationships WHERE username = ? AND datedue = ?"; 
  $tasks = $this->db->query($sql, array($username, $datedue));
  $count = 0;
      $taskdetails = array();

  foreach($tasks->result() as $row)
  {

   $taskid = $row->taskid;

    $subsql = "SELECT * FROM tasks WHERE id = ?"; 
    $taskdetails[$count] = $this->db->query($subsql, array($taskid));
    $count++;
  }

  $data['taskdetails'] = $taskdetails;
  $data['total'] = $count;

  $this->header();
  $this->load->view('dashboard', $data);

    <?php 

 foreach($taskdetails as $entry)
   {
  foreach($entry->result() as $row)
  {
   echo $row->name;
  }

   }

?>

   foreach($taskdetails as $task)
   {
     echo $task->name;
   }

---

任何帮助都很好，谢谢。

验证任务表是否包含名称字段

查看

 foreach($taskdetails as $entry)
{
   foreach($entry->result() as $row)
   {
       echo $row[0];
   }

}

为什么要在此行中传递数组

$taskdetails[$count]=$this->db->query（$subsql，array（$taskid））

而是写

$taskdetails[$count]=$this->db->query（$subsql，$taskid）

---

视图未显示某些内容的原因是您的查询不完全正确

 $datedue = 2011-01-27;

应该用引号括起来，因为日期是字符串

 $datedue = "2011-01-27";

此外，您没有正确地遵循MVC的概念。所有数据库查询和结果都应该发生在模型内部。以及控制器内部的所有数据处理

控制器和视图都应该处理任何数据库连接，这是模型的职责。因此，我建议创建一个模型，并将所有数据库查询放在一个或两个函数中。然后，控制器将调用这些函数并接收返回的数据。然后它应该操纵它，并将它以干净的方式传递给视图（我的意思是，视图永远不应该调用

result（）

以下是您的代码应该如何构造：

控制器

class Tasks extends Controller {

function Tasks ()
{
    parent::Controller();   
    $this->load->model('tasks_model');
    $this->load->database();
}

function index()
{
  $datedue = "2011-01-27";
  $username = $this->tank_auth->get_username();

  $tasks = $this->tasks_model->getTasks($username, $datedue);


  $count = count($tasks);


  $data['taskdetails'] = $tasks;
  $data['total'] = $count;

   $this->header();
  $this->load->view('dashboard', $data);
}
}

型号

class Tasks_model extends Model {


    function Tasks_model()
    {
        // Call the Model constructor
        parent::Model();
    }

    function getTasks($username, $datedue) {
        $sql = "SELECT taskid FROM relationships WHERE username = ? AND datedue = ?"; 
        $tasks = $this->db->query($sql, array($username, $datedue));

        $taskdetails = array();

        foreach($tasks->result() as $row){

            $taskid = $row->taskid;

            array_push( $taskdetails, $this->getTasksDetails( $taskid ) );
        }

        return $taskdetails;
    }

    function getTasksDetails($taskid) {

        $subsql = "SELECT * FROM tasks WHERE id = ?"; 
        $taskdetails = $this->db->query($subsql, array($taskid));

        return $taskdetails->row();


    }


}

查看：

   $datedue = 2011-01-27;
  $username = $this->tank_auth->get_username();
  $sql = "SELECT taskid FROM relationships WHERE username = ? AND datedue = ?"; 
  $tasks = $this->db->query($sql, array($username, $datedue));
  $count = 0;
      $taskdetails = array();

  foreach($tasks->result() as $row)
  {

   $taskid = $row->taskid;

    $subsql = "SELECT * FROM tasks WHERE id = ?"; 
    $taskdetails[$count] = $this->db->query($subsql, array($taskid));
    $count++;
  }

  $data['taskdetails'] = $taskdetails;
  $data['total'] = $count;

  $this->header();
  $this->load->view('dashboard', $data);

    <?php 

 foreach($taskdetails as $entry)
   {
  foreach($entry->result() as $row)
  {
   echo $row->name;
  }

   }

?>

   foreach($taskdetails as $task)
   {
     echo $task->name;
   }

---

您是否得到任何输出或完全空白的页面？我的标题显示，但不是仪表板视图尝试在foreach之外插入一些内容，如“dashboard”或其他内容是的，直到您确认“dashboard”视图实际正在加载，这很难帮助。可能会显示

$this->header（）的代码

它正在加载，我通过回显'total'变量对其进行了测试。+1，用于显示MVC的用途。此外，我还将检查他的

标题（）中发生了什么情况，这可能会阻止加载其他内容/视图。