SQL多重連接不會在Codeigniter中返回所有行

Question

這是我的查詢：

$this->db->select('e.emp_no', 'e.birth_date', 'e.first_name', 'e.last_name', 'e.gender', 'e.hire_date')
            ->from('employees AS e')
            ->join('titles AS t', 'e.emp_no = t.emp_no')
            ->join('salaries AS s', 't.emp_no = s.emp_no')
            ->join('dept_emp AS de', 's.emp_no = de.emp_no')
            ->join('departments AS d', 'de.dept_no = d.dept_no')
            ->join('dept_manager AS dm', 'd.dept_no = dm.dept_no')
            ->limit($limit, $offset)
            ->order_by($sort_by, $sort_order)
            ->group_by('e.emp_no')
            ->where('t.title', 'staff'); 

然后，我用這個返回數組：

$ret['rows'] = $q->get()->result();

        print_r ($ret['rows']);

輸出的全部是emp_no：

Array ( [0] => stdClass Object ( [emp_no] => 10002 ) [1] => stdClass Object ( [emp_no] => 10005 ) [2] => stdClass Object ( [emp_no] => 10007 ) [3] => stdClass Object ( [emp_no] => 10011 ) )

如何輸出所有選定的列？

PS：我也嘗試過使用result_array()沒有任何效果。

Answer 1

您應該將所有字段名稱組合為一個字符串，而不是將它們作為select()函數的多個參數分開

$this->db->select('e.emp_no, e.birth_date, e.first_name, e.last_name, e.gender, e.hire_date')
        ->from('employees AS e')
        ->join('titles AS t', 'e.emp_no = t.emp_no')
        ->join('salaries AS s', 't.emp_no = s.emp_no')
        ->join('dept_emp AS de', 's.emp_no = de.emp_no')
        ->join('departments AS d', 'de.dept_no = d.dept_no')
        ->join('dept_manager AS dm', 'd.dept_no = dm.dept_no')
        ->limit($limit, $offset)
        ->order_by($sort_by, $sort_order)
        ->group_by('e.emp_no')
        ->where('t.title', 'staff');