Codeigniter-克隆数据（但删除1个不存在的列） [英] Codeigniter - Clone Data (But remove 1 column that is not exist )

问题描述

我要做什么：
尝试从表付款历史到付款中克隆数据（更新），但是现在的问题是付款中历史记录表中有 PaymentHistory_ID列，我该如何忽略它？

What i am trying to do:
Try to Clone Data (Update) from Table "Payment History" to "Payment" , But the problem right now is that in Payment History Table there is "PaymentHistory_ID" column , how do I ignore it ?

付款表

Payment_ID
Payment_Amount
Payment_Method
Payment_Remark

付款历史记录表

Payment_ID
PaymentHistory_ID
Payment_Amount
Payment_Method
Payment_Remark

两者具有相同的列和相同的数据类型

Both have the same column and same data type

我的代码：
控制器

        public function updateHistory($Payment_ID){

$query = $this->db->where('Payment_ID', $Payment_ID)->get('PaymentHistory');
foreach ($query->result() as $row) {      
      $this->db->where('Payment_ID', $row->Payment_ID)->update('PaymentHistory', $row); // To update existing record
      //$this->db->insert('Payment',$row); // To insert new record
}

    }

关于我已提供
的问题 Codeigniter从表中克隆数据并更新它
，代码有效，但不适用于我面临的新问题。

Based on the question that I have provided Codeigniter Clone Data from table and update it , the codes works , but not the new problem that I am facing.

注意：CodeIgniter的新手

Note: New to CodeIgniter

推荐答案

（具有单个Payment_ID）

public function updateHistory($Payment_ID){
        // with single Payment_ID

        //get data with specific column and ignore PaymentHistory_ID
        $query = $this->db->select('Payment_Amount','Payment_Method','Payment_Remark')->where('Payment_ID', $Payment_ID)->get('YOUR FROM TABLE');
        $result = $query->row_array();

        if($result)
        { // check if has data
            //then update to another table
            $this->db->where('Payment_ID', $Payment_ID)->update('YOUR TO TABLE', $result ); // To update existing record
        }

}

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