Question

当我在codeigniter中使用此查询时：

        $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();

我遇到了这个错误

 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`

它将值1视为字段..我该如何解决？

Answer 1

将expenses.expense_type = 1移至where子句。

        $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();

编辑：更新了where子句，以也匹配与expenses连接不匹配的行。有点丑陋的解决方案，但我不太了解codeigniter的语法。检查or_where。

Answer 2

只需交换条件并在条件之间放置AND

 $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();

'on子句'代码初始化器中的未知列

2 个答案: