Question

我有3张桌子

学生

id  s_name 
1     S1
2     S2

student_subject

id  s_id  subject
1    1      english
2    1      science
3    2      mathematics
4    2      poetry

老师

id t_id s_id
1   1    1
2   1    2

我正在为老师制作一个仪表板，在那里他可以看到他下面的所有学生以及学生所关注的科目。我有一个老师的id，在控制器中被携带（如t_id），然后在模型中，从那里我从教师表中获取学生的id（如$s_id）通过这个s_id我希望从student table和student_subject表中获取详细信息。

我面临的问题是

1）在模型中，我能够看到学生表的所有细节，但是当我将值返回给控制器然后查看时我只得到1名学生的详细信息。

2）在这个模型中，除了学生的详细信息，我还希望从student_subject表返回特定学生下的科目，但我不知道如何返回2个值（即学生详细信息和科目详细信息））从1个模型到1个控制器然后再查看

我关注的代码是

控制器

public function dashboard($t_id)
        {
            $data['student_request'] = $this->student_model->student_detail($t_id);
            $this->load->view('teacher/dashboard_view',$data);
        }

模型

public function student_detail($t_id)
    {   
        $query = $this->db->query("SELECT * FROM teacher where t_id = $t_id");

        foreach ($query->result_array() as $row)
            {
                    $s_id = $row['s_id'];

                    $new_query = $this->db->query("SELECT * FROM student where id = $s_id");
                    $s = $new_query->result_array();
                    //print_r ($s); // just to check the data
                    return $s;
            }
    }

查看

<?php
foreach ($student_request as $row)
    {
        echo $row['s_name'];
    }
?>

如果有人能帮我解决问题，我真的很感激

Answer 1

不要将return放在循环中。它会立即终止循环。只需在循环外部创建一个临时容器，然后将数据推入其中，最后返回：

基本理念：

public function student_detail($t_id)
{   
    $data = array(); // initialzie container
    $query = $this->db->query("SELECT * FROM teacher where t_id = $t_id");
    $results = $query->result_array();
    foreach ($results as &$row) {

            $s_id = $row['s_id'];
            $new_query = $this->db->query("
                SELECT s.s_name, ss.subject FROM student s
                JOIN student_subject ss ON s.id = ss.s_id
                WHERE s.id = $s_id
            ");
            $s = $new_query->result_array();

            $row['student_data'] = $s; // push it here

            // return $s; // DONNT PUT RETURN HERE!
    }

    $data = $results;
    return $data;
}

或者您可以使用一个查询：

public function student_detail($t_id)
{   
    return $this->db->query("
        SELECT t.t_id, s.s_name, ss.subject 
        FROM student s
           JOIN teacher t ON t.s_id = s.id
           JOIN student_subject ss ON s.id = ss.s_id
        WHERE t.t_id = {$t_id}
        ")
        ->result_array();
}

Answer 2

修改

<强>功能

public function dashboard($t_id) { $data['student_request'] = $this->student_model->student_detail($t_id); foreach($data['student_request'] as $row) { $s_id = $row['s_id']; } $data['final_details'] = $this->student_model->get_students($s_id) $this->load->view('teacher/dashboard_view',$data); }

<强> MODEL

public function student_detail($t_id) { $this->db->select('*'); $this->db->from('teacher'); $this->db->where('t_id',$t_id); $query = $this->db->get(); return $query->result_array(); } public function get_students($s_id) { $this->db->select('*'); $this->db->from('student as s'); $this->db->join('student_subject as ss', 's.id = ss.s_id'); $this->db->where('s.id',$s_id); $query = $this->db->get(); return $query->result_array(); }

查看

您现在可以使用

foreach($final_details as $row) { echo $row['YOUR_COLUMNS']; echo $row['s_name'] . ' ' . $row['s_id'] . ' ' . $row['s_subject']; }

为方便起见，我使用了codeigniters AR。如果你不能那么多，我可以提供更多帮助

修改

好的，我已经回顾了自己的答案，发现老师是否有很多学生，只会返回最后一个学生。我会再给你一个选择。

我发现课程教师可以有多个学生，所以这里有一个代码来获取所有这些。

<强>功能

public function dashboard($t_id) { $data['student_request'] = $this->student_model->student_detail($t_id); $data['final_details'] = array(); foreach($data['student_request'] as $key => $value) { $s_id = $row['s_id']; $data['final_details'][] = $this->student_model->get_students($s_id) } $this->load->view('teacher/dashboard_view',$data); }

现在你唯一需要知道的是如何遍历那个$ key =＆gt; $值。

PHP指南有很多帮助。

Answer 3

保持控制器代码相同并更改模型和视图

模型

$this->db->select('
        teacher.s_id,
        student.s_name,
        student_subject.subject
    ');

$this->db->from('teacher');
$this->db->join('student','student.id = teacher.s_id');
$this->db->join('student_subject','student_subject.s_id = teacher.s_id');
$this->db->where('teacher.t_id',$t_id);

$query = $this->db->get();
if($query->num_rows() < 1)
    {
        return FALSE;
    }

    return $query->result();

查看

<?php if($student_request): ?>

    <?php foreach($student_request as $request): ?>

    <?php  
    echo $request->subject;
    echo $request->s_name;  

    ?>

    <?php  endforeach; ?>

    <?php endif; ?>

无法在codeigniter中从模型获取整个数据值

3 个答案: