OOP PHP give warning : mysqli_fetch_query() expect

问题:

This question already has an answer here:

• mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource or result 32 answers

First, sorry for my English.

Hi, I'm quite new to PHP, and I'm learn it in OOP. I'm doing a work project now, which use PHP.

I have created 3 PHP files, and I want to see the output from a db query.

The files are db_connect.php, login.php, and query.php.

The query.php is just for storing the query, so that if I want to write a query, I just call a function inside query.php. But when I run it, it gives me warning:

mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean given in ../query.php

I have tried to fix it, but it still gives me an error object given or null given beside this boolean given.

Any ideas what I'm doing wrong and suggestion what I suppose to do?

By the way, this is my file :

db_connect.php

<?php

class sql_con {
    public $conn;

    public function __construct()
    {
        $this_com = "localhost";
        $user = "root";
        $pass = "root";
        $db = "db";

        try {
            $this->conn = mysqli_connect($this_com,$user,$pass,$db);
            if (mysqli_connect_error())
                {
                    die("koneksi eror : ".mysqli_connect_errno()." = ".mysqli_connect_error());
                    // exit();
                }
        }
        catch(Exception $e) {
            echo $e . mysqli_connect_error();
        }
    }
}


?>

login.php

<?php
require_once('db_connect.php');
require_once("query.php");


$str_query = "select * from ";
$table     = "table";
$condition =" where user=$user and pass=$pass";

$conn   = new db_connect();
$query  = new query();
$result = $query->query_sql($conn->conn,$str_query,$condition,$table);
$result = mysqli_fetch_assoc($query);

echo $result;

$sql_c = new sql_con();
mysqli_close($sql_c->conn);

query.php

<?php
require_once("db_connect.php");

class query {
    var $query;
    var $conn;

    public function query_sql($conn, $string_query, $condition, $table)
    {
        $conn   =new sql_con();
        $query  = mysqli_query($conn,$string_query.$table.$condition);

        return $query;
    }
}

回答1:

For those who have file structure like me in OOP PHP, after some discussion in the comment above, and I tried to fix my code, and finally working (at least for me).

This my new code :

login.php

<?php
require_once('db_connect.php');
require_once("query.php");

$str_query = "select * from ";
$table     = "table";
$condition =" where user=$user and pass=$pass";

$conn   = new db_connect();
$query  = new query();
$result = $query->query_sql($conn->conn,$str_query,$condition,$table);
$row = mysqli_fetch_assoc($result);

var_dump($row);

$sql_c = new sql_con();
mysqli_close($sql_c->conn);

Query.php

require_once("db_connect.php");

class query {
    var $query;
    var $conn;

    public function query_sql($conn, $string_query, $condition, $table)
    {
        $this->conn   = $conn;
        $query        = mysqli_query($conn,$string_query.$table.$condition);

        return $query;
    }
}