PHP不显示MYSQLI结果

Question

I'm trying to create a script that i can keep referring to which would display database entries. I believe my problem is when I'm trying to display the result I have no idea on how to call for mysqli and query. the ERROR I'm getting is Call to a member function fetch_row() on a non-object in

class connectDatabase {


public $category;
public $query; 
public $mysqli;

    public function __construct($DB_HOST, $DB_USER, $DB_PWD, $DB_NAME)
    {
        $mysqli = new mysqli($DB_HOST, $DB_USER, $DB_PWD, $DB_NAME);
        return $this->mysqli = $mysqli;

    }

    public function queryInformation($category) {
        $query = "SELECT * FROM postad WHERE `category` = $this->category ORDER by date desc";
            return $this->query = $query;


    }
    public function displayResult() {
        $mysqli = $this->mysqli ; 
        $query = $this->query;
        $result = $mysqli->query($query);

        while ($row = $result->fetch_row()) {
        echo "$row[1] $row[2] $row[3] $row[4]  </br>" ;

    }

    }
}

Answer 1

if you var_dump $result it's probably boolean false because the query is failing. It should be an object if it was successful. You need to have a fail safe in case the return value is boolean.

Check out the return values: http://php.net/manual/en/mysqli.query.php

Answer 2

We can simplify this a bit.

Try this

    $mysqli = new mysqli($DB_HOST, $DB_USER, $DB_PWD, $DB_NAME);
    $category = "something";

    $result = $mysqli->query("SELECT * FROM postad WHERE `category` = '$category' ORDER by date desc");
    while ($row = $result->fetch_row()) {
        echo "$row[1] $row[2] $row[3] $row[4]  </br>" ;

    }