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mysql - “on 子句”codeigniter 中的未知列

问题描述

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();

我收到了这个错误

 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？

标签： mysqlcodeigniter

解决方案

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();

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