这是我的连接文件
<?php
$mysqli= @new mysqli("localhost","root","","ngo_sharda");
if($mysqli->connect_errno)
{
printf ("connection failed %s 'n",mysqli_connect_error());
}
?>
这是我的班级文件...
<?php
include('../connection.php');
class operation
{
private $title;
/*
function __construct()
{
$this->title=$m;
}
*/
function setvalues()
{
$this->title=$m;
}
function insert()
{
$q="insert into menus(title,link) values('kumar','great')";
//$result=mysqli_query($mysqli,$q);
$result= $mysqli->query($q);
if($result==1)
{
echo "inserted";
}
else
{
echo "not inserted";
}
}
}
?>
如果我尝试在类中创建插入函数，则会收到错误，指出我正在对非对象调用查询。
如何在插入函数中直接在此类中调用$mysqli对象，而无需将其作为任何函数或构造函数中的参数传递。
我会在
构造函数中加载数据库连接，以便您可以使其成为类的成员。
<?php
class operation{
private $title;
private $mysqli;
function __construct(){
// include basically copies & pastes the file
include('../connection.php');
// $mysqli exists inside this function, let's make it
// available in the rest of the class
$this->mysqli = $mysqli;
//$this->title = $m;
}
function setvalues(){
$this->title = $m;
}
function insert(){
$q = "insert into menus(title,link) values('kumar','great')";
// Now we can access `$this->mysqli` and everything should work
$result = $this->mysqli->query($q);
if($result==1){
echo "inserted";
}
else{
echo "not inserted";
}
}
}
?>