PHP not displaying MYSQLI results

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>" ;

    }

    }
}

Why are you referencing $this->category when you are supposedly passing it as a parameter? — lc.
– lc., Commented Jan 28, 2013 at 4:14
what is your problem exactly? i don't feel like you addressed the problem you're having. — Emery King
– Emery King, Commented Jan 28, 2013 at 4:17

Emery King · Accepted Answer · 2013-01-28 04:32:48Z

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

answered Jan 28, 2013 at 4:32

Emery King

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OrganizedChaos · Accepted Answer · 2013-01-28 04:17:25Z

0

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>" ;

    }

answered Jan 28, 2013 at 4:17

OrganizedChaos

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4 Comments

Emery King Over a year ago

that's the same thing without the re usability of oop.

Jan Schejbal Over a year ago

@MarshallHouse: Which isn't a bad thing for a code example to find the error.

Anthony Liriano Over a year ago

I'm trying to implement OOP. If I wasn't it'd be a lot easier. The error I'm gettting is Fatal error: Call to a member function fetch_row() on a non-object in The error should be in while ($row = $result->fetch_row()) or $result = $mysqli->query($query);

OrganizedChaos Over a year ago

@AnthonyLiriano Gotya, didn't realize OOP was the goal. I was just trying to simplify it.

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