Question

型号：

function get_all() {
        $query = $this->db->get('loan');
        return $query->result();
    }

    function get_loan_name($code) {
        $this->db->where('loan_type_code',$code);
        $query = $this->db->get("loan_type");
        return $query->row();
    }

控制器：

function index()
{
    $this->data['loans'] = $this->loan_approval->get_all();
    $this->build_view("loan_approval/list.php");
}

在“LOAN TABLE”中，它有一个“loan_type_code”列，我希望获得此列中的所有值并匹配第二个函数。实施例

Loan table                             Loan Type Table

ID     loan_type           ID     loan_type_code   loan_type_name

1         EDL              1           EDL          Education Loan

2         DF               2           DF            Damayan Fund

我需要将表“LOAN”中的“loan_type”与“Loan Type”表中的“loan_type_code”相匹配。我想是否像这样的查询“$ this-＆gt; controler-＆gt; get_loan_name（$ this-＆gt; data ['loans'] - ＆gt; loan_type）;”但它不起作用。

Answer 1

为什么不使用JOIN在数据库级别执行此操作？

$query = $this->db->select('loan.*, loan_type.loan_type_name')
                  ->from('loan')
                  ->join('loan_type','loan.loan_type = loan_type.loan_type_code')
                  ->get();
return $query->result();

如何使用result（）获取一列中的所有值

1 个答案: