【问题标题】:Codeigniter: You must use the "set" method to update an entryCodeigniter:您必须使用“set”方法来更新条目
【发布时间】:2016-06-24 18:08:56
【问题描述】:

我正在尝试使用提交的表单插入数据。数据在数据库中正确输入,但我也在为移动设备做 API。如果收到is_api 参数,我正在做jsone_encode。以下是代码:

控制器:

public function send_request(){
            if($this->input->post('submit')){        
                $user_data['data'] = array( 
                    'user_id'                =>  1,
                    'sender_name'            =>     $this->input->post('sender_name'),
                    'sender_location'        =>     $this->input->post('sender_location'),
                    'sender_mobile'          =>     $this->input->post('sender_mobile'),
                    'receiver_name'          =>     $this->input->post('reciever_name'),
                    'receiver_location'      =>     $this->input->post('reciver_location'),
                    'receiver_mobile'        =>     $this->input->post('reciver_location'),
                    'request_type'           =>     $this->input->post('request_type'),
                    'is_urget'              =>     0,
                );

            }
            $result = $this->user_model->send_request($user_data);
             if($result){
                $this->result_set['message'] = 'Data entered ';
                $this->result_set['status']= 1;
             }
                if( $this->input->post('is_api') == 1){
                    echo json_encode($this->result_set);
                    die();
                }
         }

型号:

public function send_request($data){
        $this->db->insert('parcel_requests',$data['data']);
        return true;
     }

当我检查 hurl.it 的 API 时,我收到此错误和响应:

发生数据库错误

您必须使用“set”方法来更新条目。

文件名:/home/foldername/public_html/parcel/models/user_model.php

行号:21

这是第 21 行:

    $this->db->insert('parcel_requests',$data['data']);

现在数据已正确输入,但未转换为 json。我需要你的帮助来解决这个问题,谢谢!

【问题讨论】:

  • 您是否尝试仅将 $data 数组而不是 $data['data'] 传递给插入函数?
  • 用$sql = $this->db->set($data['data'])->get_compiled_insert('parcel_requests'); echo $sql;检查你的查询字符串

标签: php mysql codeigniter api


【解决方案1】:

检查一下:

public function send_request(){
    if($this->input->post('submit')){
        $user_data['data'] = array(
            'user_id'                =>  1,
            'sender_name'            =>     $this->input->post('sender_name'),
            'sender_location'        =>     $this->input->post('sender_location'),
            'sender_mobile'          =>     $this->input->post('sender_mobile'),
            'receiver_name'          =>     $this->input->post('reciever_name'),
            'receiver_location'      =>     $this->input->post('reciver_location'),
            'receiver_mobile'        =>     $this->input->post('reciver_location'),
            'request_type'           =>     $this->input->post('request_type'),
            'is_urget'              =>     0,
        );
        // inside your if ;), you should insert to database only if you submit data... 

        $result = $this->user_model->send_request($user_data);
        if($result){
            $this->result_set['message'] = 'Data entered ';
            $this->result_set['status']= 1;
        }
        if( $this->input->post('is_api') == 1){
            echo json_encode($this->result_set);
            die();
        }
    } else {
        print 'nthg sumbited';
    }
 }

public function send_request($data){
    $this->db->insert('parcel_requests',$data['data']);
    return $this->db->affected_rows() == 1;
    // Will return true after insert will be ended with success
}

在您的版本中,有可能发送空数组:),这就是显示此异常的原因:)

【讨论】:

    【解决方案2】:

    将您的 $user_data['data'] = array 更改为 $data['data'] = array

    【讨论】:

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