php - CodeIgniter:如何将 $this->db->insert_id() 值传递给 Controller 以便插入到另一个表中？

Question

我正在尝试向表中插入字段并获取 ID(假设表的主键是 AUTO INCREMENT)。我需要将该记录的自动增量 ID 插入到第二个表中。我显然可以返回 $this->db->insert_id() 但我如何在之后访问 Controller 中的值？

我尝试在变量中定义它并在 Controller 中访问，但没有成功。

A PHP Error was encountered
Severity: Notice
Message: Undefined variable: insert_id
Filename: controllers/POS.php
Line Number: 87

型号:

function populate_pos_purchase_table($posPurchase) {
    $this->db->trans_begin();
    $this->db->insert('pos_purchase', $posPurchase);

    if ($this->db->trans_status() === FALSE) {
        $this->db->trans_rollback();
        return false;
    }
    else {
        $this->db->trans_commit();
        $insert_id = $this->db->insert_id();
        return $insert_id;
        return true;
    }
}

Controller :

function update_payment_details() {

$newPayment = array (
    'pos_id' => $insert_id, // Line Number: 87
    'payment_description' => 'Point of Sales',
    'payment_method' => $this->input->post('payment_method'),
    'payment_date' => $this->input->post('payment_date'),
    'paid_amount' => $this->input->post('total'),
    'due_amount' => 0
);

$this->pos_model->populate_new_payment_table($newPayment);  
$this->index();

} 

Answer 1

只需保存第一个插入的 ID - 然后在第二个插入中使用它？

function update_payment_details() {

$insert_id = populate_pos_purchase_table($posPurchase);

$newPayment = array (
    'pos_id' => $insert_id, // Line Number: 87
    'payment_description' => 'Point of Sales',
    'payment_method' => $this->input->post('payment_method'),
    'payment_date' => $this->input->post('payment_date'),
    'paid_amount' => $this->input->post('total'),
    'due_amount' => 0
);

$this->pos_model->populate_new_payment_table($newPayment);  
$this->index();

}