使用codeigniter连接表并将数据提取到表中

Question

我有两个名为verification_details和的数据库表verification_questions. verification_id在两个表格中很常见.在verification_details表格中,有一个user_id字段,基于该user_id字段,插入verification_details表格中的一个或多个验证细节,并基于插入表格中的verification_id一个或多个验证问题verification_questions.我在模型中使用此代码进行连接查询

function get_records() {

    $this->db->select("a.criteria_question,a.criteria_answer,a.validation_statement");
        $this->db->from("verification_questions as a");
        $this->db->join('verification_details as b', 'a.verification_id = b.verification_id');
        $query = $this->db->get();
        return $query->result();    
}

我想要控制器代码从两个表中获取所有数据并将其显示在表中.

Answer 1

在模型中

function get_records() {

    $this->db->select("a.criteria_question,a.criteria_answer,a.validation_statement");
    $this->db->from("verification_questions as a");
    $this->db->join('verification_details as b', 'a.verification_id = b.verification_id');
    $query = $this->db->get();
    $result = $query->result_array();
    return $result;
}

在控制器中

$data['table'] = $this->Model_name->get_records();
$this->load->view('view_name',$data);

在视野中

<table>
    <tr>
        <th>Question</th>
        <th>Answer</th>
        <th>Validation</th>
    </tr>
    <?php
        foreach ( $table as $new_item )
        {
            ?>
            <tr>
                <td><?php echo $new_item['table field1']?></td>
                <td><?php echo $new_item['table field2']?></td>
                <td><?php echo $new_item['table field3']?></td>
            </tr>
        <?php
        }

    ?>
</table>