Неизвестный столбец в коде «on clause»

Question

когда я использую этот запрос в codeigniter:

        $this->db->where("dopuex_donation_id", $donation_id);
        $this->db->select("donation_purposes_expenses.*, SUM(donation_purposes_expenses.dopuex_withdrawn_value) AS count, cases.case_name");
        $this->db->from("donation_purposes_expenses");

        $this->db->join("expenses", "donation_purposes_expenses.dopuex_expense_id = expenses.expense_id", "left");
       $this->db->join("cases", "expenses.expense_type = 1 && expenses.expense_case_ref_id = cases.case_ref_id", "left");

        $this->db->group_by("cases.case_name");
        return $this->db->get()->result_array();

я получил эту ошибку

 Unknown column '1' in 'on clause'

SELECT `donation_purposes_expenses`.*, SUM(donation_purposes_expenses.dopuex_withdrawn_value) AS count, `cases`.`case_name` FROM (`donation_purposes_expenses`) LEFT JOIN `expenses` ON `donation_purposes_expenses`.`dopuex_expense_id` = `expenses`.`expense_id` LEFT JOIN `cases` ON `expenses`.`expense_type` = `1` && expenses.expense_case_ref_id = cases.case_ref_id WHERE `dopuex_donation_id` = '34' GROUP BY `cases`.`case_name`

он рассматривает значение 1 как поле .. как я могу это исправить?

Answer 1

Просто поменяйте местами условия и поместите И между условиями

 $this->db->where("dopuex_donation_id", $donation_id);
     $this->db->select("donation_purposes_expenses.*, SUM(donation_purposes_expenses.dopuex_withdrawn_value) AS count,
     cases.case_name");
    $this->db->from("donation_purposes_expenses");
    $this->db->join("expenses", "donation_purposes_expenses.dopuex_expense_id = expenses.expense_id",
     "left");
    $this->db->join("cases", "expenses.expense_case_ref_id = cases.case_ref_id AND  expenses.expense_type = 1 ", "left");
    $this->db->group_by("cases.case_name");
    return $this->db->get()->result_array();

Answer 2

Переместите expenses.expense_type = 1 к предложению where.

        $this->db->where("dopuex_donation_id", $donation_id);
        $this->db->where("expenses.expense_type is NULL OR expenses.expense_type = 1", NULL, false); // New where clause !!
        $this->db->select("donation_purposes_expenses.*, SUM(donation_purposes_expenses.dopuex_withdrawn_value) AS count, cases.case_name");
        $this->db->from("donation_purposes_expenses");

        $this->db->join("expenses", "donation_purposes_expenses.dopuex_expense_id = expenses.expense_id", "left");
        $this->db->join("cases", "expenses.expense_case_ref_id = cases.case_ref_id", "left");

        $this->db->group_by("cases.case_name");
        return $this->db->get()->result_array();

Редактировать: Обновлено предложение where, чтобы соответствовать также строкам, которые не соответствуют соединению expenses.Вид некрасивого решения, но я не очень хорошо знаю синтаксис codeigniter.Проверьте или где.