Question

在提交表单时，我在codeigniter中收到此错误

severity: Warning

Message: mysqli_query() expects at least 2 parameters, 1 given

Filename: models/Login_model.php

Line Number: 50

Backtrace:

File: /opt/lampp/htdocs/zemco/application/models/Login_model.php
Line: 50
Function: mysqli_query

File: /opt/lampp/htdocs/zemco/application/controllers/Stock_main.php
Line: 42
Function: addCustomerDb

File: /opt/lampp/htdocs/zemco/index.php
Line: 292

功能：require_once

控制器：

public function __construct()
{
      parent::__construct();
      $this->load->library('session');
      $this->load->helper('form');
      $this->load->helper('url');
      $this->load->helper('html');
      $this->load->database();
      //load the login model
      $this->load->model('Login_model');
}

public function addCustomer(){
    $result = $this->Login_model->addCustomerDb();
}

型号：

public function addCustomerDb(){

    echo "INSERT INTO `tbl_customer` (customer_company_name, customer_primary_address1, customer_primary_address2, customer_primary_country, customer_primary_state, customer_primary_city, customer_primary_poboxno, customer_secondary_address1, customer_secondary_address2, customer_secondary_country, customer_secondary_state, customer_secondary_city, customer_secondary_poboxno, customer_landline, customer_fax, customer_internal_notes, customer_invoice_flag, customer_created_time, customer_created_by) VALUES('$comName', '$primaryAddressstreet1', '$primaryAddressstreet2', '$country1', '$state1', '$cityName1', '$poibox1', '$secondaryAddressstreet1', '$secondaryAddressstreet2', '$country2', '$state2', '$cityName2', '$poibox2', '$businessPhone', '$faxNumber', '$internalNotes', '$emailvia', '$createdOn', '1')";
    $query = mysqli_query("INSERT INTO `tbl_customer` (customer_company_name, customer_primary_address1, customer_primary_address2, customer_primary_country, customer_primary_state, customer_primary_city, customer_primary_poboxno, customer_secondary_address1, customer_secondary_address2, customer_secondary_country, customer_secondary_state, customer_secondary_city, customer_secondary_poboxno, customer_landline, customer_fax, customer_internal_notes, customer_invoice_flag, customer_created_time, customer_created_by) VALUES('$comName', '$primaryAddressstreet1', '$primaryAddressstreet2', '$country1', '$state1', '$cityName1', '$poibox1', '$secondaryAddressstreet1', '$secondaryAddressstreet2', '$country2', '$state2', '$cityName2', '$poibox2', '$businessPhone', '$faxNumber', '$internalNotes', '$emailvia', '$createdOn', '1')");
    $cid = $this->db->insert_id();
    echo $cid;

    $count = $this->input->post('contactCount');

    for($i = 1; $i <= $count; $i++){
        if($this->input->post('customerEmail1'.$i) != ''){
            $customerEmail = $this->input->post('customerEmail1'.$i);
            $customerPhone = $this->input->post('customerPhoneNumber'.$i);
            $customerMobile = $this->input->post('customerMobileNumber'.$i);
            $firstName = $this->input->post('firstName'.$i);
            $lastName = $this->input->post('lastName'.$i);

            $contAdd = mysqli_query("INSERT INTO `tbl_customer_contact` (customer_id, customer_contact_name, customer_contact_email, customer_contact_mobile, customer_contact_phone) VALUES('$cid', '$firstName"."$lastName', '$customerEmail', '$customerMobile', '$customerPhone')");
        }
    }

    if($query){
      //return 1;
    }
}

我在codeigniter中执行表单发布时遇到此错误。我不知道我在哪里滞后。

Answer 1

我犯了小错误，谢谢你的回复。我试过了，

$query = mysqli_query("");

替换为

$query = $this->db->query("");//i used this code

我再一次感谢你们。

Answer 2

If you use mysqli_query as a function and not an method from a mysqli-object you have to pass the mysqli-link as the first parameter.

php.net Manual mysqli.query

mysqli_query（）需要至少2个参数，1在codeigniter中给出

2 个答案: