Question

Warning: mysqli_query() expects parameter 1 to be mysqli, object given in

class mysqlConn extends sqlCrud{
private $conn;
private $table;
private $server;
private $username;
private $password;
public function mysqlConn($server,$username,$password,$table){
$this->conn=mysqli_connect("$this->server","$this->username","$this->password",$this->table);
if(!$this->conn){
die(mysql_error());
}
}
}
$mysqlconn=new mysqlConn("localhost","root","123456","test");
$sql="SELECT * FROM test";
$query=mysqli_query($mysqlconn,$sql) or die(mysql_error());
最后一句为什么错了？

Answer 1

mysql_query();你这里面的参数写反了吧，这是例子：

<?php
$con = mysql_connect("localhost","mysql_user","mysql_pwd");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

$sql = "SELECT * FROM Person";
mysql_query($sql,$con);

// 一些代码

mysql_close($con);
?>

Answer 2

改成这样

$query=mysqli_query($mysqlconn->conn,$sql) or die(mysql_error());

追问

conn是private，不能访问的

追答

没注意，那你在类里写个public的方法调用就好了

追问

我自己解决了