Question

我想将连接链接传递给一个类。我尝试这样但无法得到解决方案这里 dbconnect.php

<?php
global $con;
$db_host="localhost";
    $db_username="root";
    $db_pass="";
    $db_name="xxx";

$con= mysqli_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql"); 

mysqli_select_db($con,$db_name) or die ("no database"); 

?>

和我的类php文件

<?php
include('dbconnect.php');
$chechout=new clscheckout;


$chechout->con=$con;  // direct assign connection link
$chechout->adult_count=2;
$chechout->child_count=1;
$chechout->child_cwb_count=1;
$chechout->setConn();   // assign throug function
echo $chechout->fnTotel();


class clscheckout{
   public  $pack_id;
   public  $pack_title;
   public  $user_id;
   public  $adult_count;
   public  $child_count;
   public  $child_cwb_count;
   public  $con;


      function setConn ($conn) {
          $con=$conn;
      }

       function fnTotel () {
            $qry='select * from package where pack_id='.$pack_id ;
            $rs=mysqli_query($con,$qry);
            if($rs)
                {
                $row=mysqli_fetch_array($rs);
                $child_wob_price=$row['child_wob_price'];
                $adult_price=$row['adult_price'];
                $child_price=$row['child_price'];
                }

            $adult_total=$adult_price*$adult_count;

            $chidl_total=$child_price*$child_count;
            $child_wob_total=$child_wob_price*$child_cwb_count;

            return $adult_total+$chidl_total+$child_wob_total;

       }

}

?>

我直接和通过函数分配了类似的东西。但没有成功

Answer 1

首先，我将通过构造函数

分配连接

class clscheckout {

    // snip

    private $con;

    public function __contruct(mysqli $con) {
        $this->con = $con;
    }

    // etc

并使用

实例化它

require_once 'dbconnect.php';
$chechout = new clscheckout($con);

然后，请确保在$this->con方法中使用clscheckout而不是$con。这适用于所有类属性。见http://www.php.net/manual/language.oop5.properties.php

您的global $con脚本中也不需要dbconnect.php。

Answer 2

最好的方法是先创建一个mysqli类的实例：

dbconnect.php

$mysqli = new mysqli($host, $user, $pass, $db);

if($mysqli->connect_error) {
    die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error());
}

return $mysqli;

然后

$connection = require_once('dbconnect.php');
$chechout = new clscheckout($connection);

并按照您的意愿操纵/传递它。这样你就不会使用讨厌的全局

class clscheckout{
   public  $con;


      public function __construct ($conn) {
          $this->con=$conn;
      }

然后将其用作对象

$qry='select * from package where pack_id='.$pack_id ;
$rs=$this->con->query($qry);

Answer 3

您是否尝试过此操作（通过引用传递$ conn而不是按值传递）？

function setConn (&$conn) {
      $con=$conn;
}

有关更多信息，请参阅http://php.net/manual/en/language.references.php。

如何将数据库连接传递给类

3 个答案: