Question

我正在尝试向表中插入字段并获取id（假设表的PRIMARY KEY是AUTO INCREMENT）。我需要将该记录的自动增量ID插入第二个表中。我显然可以返回$ this-＆gt; db-＆gt; insert_id（），但之后如何访问Controller中的值？

我尝试在变量中定义它并在控制器中进行访问，但它不起作用。

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;
    }
}

控制器：

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吗？

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();

}

CodeIgniter：如何将$ this-＆gt; db-＆gt; insert_id（）值传递给Controller以便插入另一个表中？

1 个答案: