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在我的代码中，我只能搜索一个表并从一个表中返回一些东西，但是从发布的值姓氏中，我如何获取“emp_no”以便我可以从“dept_emp”返回“dept_no”和“title”从“头衔”？

控制器

public function findemp() 
{   
    // search in table 'employees'
    $name = $this->input->get('lastname');


    if($data['user'] = $this->fndmodel->finduser($name)) {

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

    // search in table 'dept_emp'
    $dept = $this->input->get('deptmnt');
    // search in table 'titles'
    $title = $this->input->get('title');
}

该模型

 public function finduser($name) {

    if ($name == null || $name == '') {
            return false;
    }

    $this->db->select('e.emp_no, e.first_name, e.last_name, t.title, d.dept_no');
    $this->db->from('employees AS e, titles AS t, dept_emp AS d');
    $this->db->where('e.emp_no = t.emp_no');
    $this->db->where('e.emp_no = d.emp_no');
    $this->db->where('e.last_name', $name);
    $data['query'] = $this->db->get();              
    return $data;

}

这就是我想要的 activerecords

SELECT e.emp_no, e.first_name, e.last_name, t.title, d.dept_no
FROM employees e JOIN titles t
ON (e.emp_no = t.emp_no)
JOIN dept_emp d
ON (e.emp_no = d.emp_no)
WHERE e.last_name = "facello";

这是正确的吗？

$this->db->select('e.emp_no, e.first_name, e.last_name, t.title, d.dept_no');
$this->db->from('employees AS e, titles AS t, dept_emp AS d');
$this->db->where('e.emp_no = t.emp_no');
$this->db->where('e.emp_no = d.emp_no');
$this->db->where('e.last_name', $name);
$result = $this->db->get();
4

0 回答 0